Solving Quadratic Trigonometric Equations Quadratic trigonometric equations are ones that involve products of trigonometric functions.
EXAMPLE 1: Solve the trigonometric equation 2cosx+2=3. A). Solve for cos x – cos x = ½ B). Solve for x by finding all the values in the interval 0 < x < 360 that satisfy the equation. In this case, with cosine being positive and equal to ½ , there are two values. One in the first quadrant of the unit circle – x = pi/3. C). Therefore, the second one is in the forth quadrant since cosine is positive there too. Making pi/3 negative and adding 360, your second value is 5pi/3. D). Your answers are 60 degrees and 300 degrees.
EXAMPLE 2: Solve tan squared x-3=0. A). Add 3 to the right side of the equal sign. B). Since tan is squared, find the square root of tan squared and 3. – tan x = +/- square root of 3 C). Your answers are 60 degrees and 120 degrees. *When dealing with a trig function that has a power, it will be solved by extracting square roots or factoring.
EXAMPLE 3: Solve 2sin squared x + sin x -1 = 0. A). Since we’re dealing with just sin, you can factor. B). (2sinx-1)(sinx+1) = 0 C). 2sinx-1 = 0 / sinx+1 = 0 D). sin x = ½ / sin x = -1 E). x = 30 degrees, 150 degrees and x = 270 degrees
**Note that many of the techniques used in solving algebraic equations are also used to solve trigonometric equations.
There are a few different ways that you can solve a quadratic formula. This is something that we learned in algebra 2, but it takes a bit of practice. Sometimes factoring can be used to solve a quadratic, but the quadratic formula will ALWAYS work every time. The formula is x = - b +/- square root (b^2- 4ac)/ 2a
Example 1: Solve x^2 + 3x – 4 = 0 a. Ths problem could be solved by factoring, but using the quadratic formula, you just plug it in. x = -3 +/- square root( (3) ^2 -4(1)(-4)/ 2(1) b. Now, that it is plugged in, you just solve. -3 +/- square root(9 + 16)/ 2 c. -3 +/- square root(25)/2 d. -3 +/- 5/2 e. -3-5/2 and -3 + 5/ 2 ( Simply make two equations. ) f. -8/2 and 2/2 = -4, 1
x= -4 and x= 1
Example 2: x(x-2) =4
In order to use the quadratic formula, you must have everything equal to zero, and that is not the case in this problem. a. x(x-2)= 4 b. x^2 -2x = 4 c. x^2 -2x -4 = 0 Now you can plug it into the formula. a. x = -(-2) +/- square root( (-2) ^2 – 4(1)(-4) ) / 2(1) b. 2 +/- square root(4 +16)/2 c. 2 +/- square root(20)/ 2 d. 2 +/- square root(4) square root(5)/ 2 e. 2 +/- 2 square root(5)/2 f. 2(1 +/- square root(5)/ 2(1) = 1 +/- square root(5) g. You will get decimal answers, but you round them two decimal places.
A trigonometric equation is one that involves one or more of the six functions sine, cosine, tangent, cotangent, secant, and cosecant. Some trigonometric equations, like x = cos x, can be solved only numerically. But a great many can be solved in closed form, and this page shows you how to do it in five steps. Contents: Overview Step 1. Get one function of one angle. Step 2. Solve for the value(s) of a trig function. Step 3. Solve for the angle. Step 4. Solve for the variable. Step 5. Apply any restrictions. More Examples What’s New Copying: You’re welcome to print copies of this page for your own use, and to link from your own Web pages to this page. But please don’t make any electronic copies and publish them on your Web page or elsewhere. Overview
If a trig equation can be solved analytically, these steps will do it: Put the equation in terms of one function of one angle. Write the equation as one trig function of an angle equals a constant. Write down the possible value(s) for the angle. If necessary, solve for the variable. Apply any restrictions on the solution.
http://oakroadsystems.com/math/trigsol.htm
When the trig function has a power, it will have to be solved by extracting square roots or factoring.
Example 1:
Solution:
Example 2:
Solution:
At this point we know that :
Now, implies that (see graph). Since the sine function has maximum and minimum values of +1 and -1, has no solutions. Thus the answer is the only solution.
There are many ways to solve quadratic trig equations. The two main ways are completing the square and using the quadratic formula. I’m going to show you how to use the quadratic formula because we like to use formulas. The quadratic formula is something we learned in Algebra 2. The formula: x = -b +/- square root (b2 – 4ac)/2a
There are a few different ways that you can solve a quadratic formula. We learned dealing with quadratic formula in algebra 2. Quadratic can be factored to solve. Quadratic formula will never fail for u. It will work every time.
Formula is the formula is x = - b +/- square root (b^2- 4ac)/ 2a.
Examples of Quadratic Formula:
Solve tan squared x-3=0. A). Add 3 to the right side of the equal sign. B). tan is squared; find the square root of tan squared and 3. – Tan x = +/- square root of 3 C). Your answers are 60 degrees and 120 degrees.
a = 1, b = –4, and c = –8.
X= -(-4) +/- square root of -4squared-4(1)(-8)/2(1)
A trigonometric equation is one that involves one or more of the six functions sine, cosine, tangent, cotangent, secant, and cosecant. Some trigonometric equations, like x = cos x, can be solved only numerically. But a great many can be solved in closed form, and this page shows you how to do it in five steps.
EXAMPLE A: 3tan²(B/2) − 1 = 0
1.To solve it, add 1 to both sides and divide by 3: tan²(B/2) = 1/3
2.Then take square root of both sides:
Final answers: tan(B/2) = ±√(1/3) or ±√(3)/3
Be sure to remember to use the plus-or-minus sign ± when taking the square root of both sides; if you do not there may be some solutions you have overlooked because of it.
EXAMPLE B: 2 cos x + 2 = 3
1.solve for cos(x) cos x = 1/2
2.solve for x by finding all values in the interval [0 , 2pi) that statisfy the above trigonometric equation.
3.In this case, with cosine positive and equal to 1 / 2, there are two values: one in the first quadrant of the unit circle. x1 = pi / 3 and a second one in the fourth quadrant (see the two solutions in unit circle in figure below).
I actually remember something about quadratic equations from algebra 2! There are 2 main ways, the quadratic formula and completing a square. Both are fairly easy.
FORMULA: -b +/- square root b^2-4ac/2a
EXAMPLE: x^2+2x-3=0 x=-2+/- square root of 2^2 -(4)(1)(-8)/2(1) this is very simple because now you just solve the equation. -2+/- (4)+(12)/2 -2+/- square root of 16/2 -2+4/2 and -2-4/2 when solving for a quadratic you must have 2 equations. -2+2 and -2-2 x=0 and x=-4
VERY SIMPLE! since you have a formula you easily just plug it in. As long as you know this quadratic formula you're fine.
Trigonometry is ridiculous. There are so many ways to do the same thing. This is just another way to do just that. Solve sin2(x) – sin(x) = 2 on 0° < x < 360° This is a quadratic in sine, so I can apply some of the same methods: sin2(x) – sin(x) – 2 = 0 (sin(x) – 2)(sin(x) + 1) = 0 sin(x) = 2 (not possible!) or sin(x) = –1 Only one of the factor solutions is sensible. For sin(x) = –1, I get: x = 270° Solve cos2(x) + cos(x) = sin2(x) on 0° < x < 360° I can use a trig identity to get a quadratic in cosine: cos2(x) + cos(x) = sin2(x) cos2(x) + cos(x) = 1 – cos2(x) 2cos2(x) + cos(x) – 1 = 0 (2cos(x) – 1)(cos(x) + 1) = 0 cos(x) = 1/2 or cos(x) = –1 The first trig equation, cos(x) = 1/2, gives me x = 60° and x = 300°. The second equation gives me x = 180°. So my complete solution is: x = 60°, 180°, 300°
There isn't one way to solve a quadratic formula but the main way to solve it is the quadratic formula. Another way to do it would be to factor in the simplest situation ax^2 + bx + c and then factor. You can also use the complete the square method. But the way that it will always work is to use the quadratic formula. The quadratic formula says that x= -b +/- the square root of (b^2-4ac)/2a. You plug the numbers into the formula and come up with the answer. The key is to know the formula, which is also the key to Chapter 10(:.
Example 1: 4x^2 + 4x -2 = 0
a. Plug the correct numbers into the quadratic formula. x= -4 +/- the square root of (4^2- 4(4)(-2)/2(4)
b.Multiply everything you can. x= -4 +/- the square root of (16+32)
c. complete the problem x= -4 +/- the square root of (48)/2
d. There are 2 answers a negative and a positive.
Example 2:
a. 3x^2+5x-2=0
b. x=-5 +/- the square root of (5^2-4(3)(-2)/2
c. 5 +/-the square root of (25+24)/2
d.5 +/- the square root (49)/2
e.(5+7)/2= 12/2=6
f. (5-7)/2= (-2/2)= -1
g. there are two answers in this problem because it is + or - x=6 and -1
In section ten four we did solved quadratic trig equations . The good thing about solving them is that you solve them the exact way you would solve a regular quadratic equation. Although the quadratic formula method may be used , the factoring method is more simple. Here are some examples : 3sin^2x+sinx=-2 In order to solve a quadratic you must have the equation equal to zero. In order to do so you must bring the negative two over to get: 3sin^2x+sinx+2=0 Now you would solve as if you would a regular quadratic equation. Here is an example of solving regular quadratic equations. You must first know how to solve a regular quadratic equation before you will be able to solve one with a trig function. Example number one : very simple.. 3x^2 + 12x=-9 first bring th negative nine over.. 3x^2+12x+9=0 next step is to factor.. factors of 3(9) that add to = 12; (9,3) 3x^2+3x+9x+9=0 factor out.. 3x(x+1)+9(x+1) now u solve.. set = to 0 3x+9=0 x+1=0 X=-3 x=-1
Example number two: 3sin^2x+sinx+2=0 =3 sin^2x +3sinx+2sinx+2=0 the next step is to factor.. =3sinx(sinx+1)2(sinx+1) set both equations equal to zero.. (3sinx+2)=0 (sinx+1)=0 now you will solve.. Sinx=3/2 sin x=1 X=sin-1(3/2) x=sin-1(1) Find theinverses and you have your answers.. Simple enough :)
When solving a quadratic for a trig equation you go back to the beginning ways of solving a quadratic, which is what you learn in Algebra 2. You can either:
1.Complete the square
Solve x2 - 6x + 2 = 0 by completing the square x2 - 6x = -2 [To complete the square on the LHS (left hand side), we must add 62/4 = 9. We must, of course, do this to the RHS also]. ∴ x2 - 6x + 9 = 7 ∴ (x - 3)2 = 7 [Now take the square root of each side] ∴ x - 3 = ±2.646 (the square root of 7 is +2.646 or -2.646) ∴ x = 5.646 or 0.354
2.Use the quadratic formula
ax2 + bx + c = 0 ---> x= -b± √( b2-4ac)/2a
3.Factor
Solve x2 + 2x - 8 = 0 ∴ (x - 2)(x + 4) = 0 ∴ either x - 2 = 0 or x + 4 = 0 ∴ x = 2 or x = - 4
When given in the form of a trig equation you do the same..
1. Some of these are copied from each other. You will not get credit if you have copied any part of some else's blog.
2. Many of you forgot there are two angles for each inverse with exceptions at 1 and -1. sin^-1 (1/2) has two quadrants where sin is positive.
3. If you gave examples without trig functions redo them with trig functions in a new comment to get credit.
4. We NEVER used the quadratic formula to solve a trig equation. You need to revisit your notes on these. This is why you are having trouble with 8-4 & 10-4. Relook at the example problems from those sections.
This week we learned how solve quadratic trig equation is section 10-4. Section 10-4 is very similar to section 8-4 but the only difference is that 10-4 involves all 8-4 formulas when you need to plug in to replace and all the formulas in all of chapter 10. Remember there are six functions in a trigonometry equation.
The six functions are: Sine Cosine Tanganent Cotangent Cosecant Secant
Examples:
Cos2x=1-sinx for 0< equal to x< equal to 360 degrees
-2sinx^2+sinx=0 Sinx(-2sinx+1)=0
Sinx=0 X=sin inverse of 0 Equal to 0,180 degrees, 360 degrees=answer also
Sinx=1/2 X=sin invere(1/2) Equal to 30 degrees where sine is + 1st quadrant and 2nd quadrant 180-30=150 Answers are 150 degrees, 30 degrees
3cos2x+cosx=2 for 0<equal to x<equal to 360 degrees 3(cos^2x-1)+cosx+/-2 6cos^2x-3+cosx=2 6 cos^2x+ cosx-5=0 Cosx(6cosx+1)-5=0 Cosx(6cosx+1)=5
Cosx=0 X=cos inverse of 0 =90 degrees Answers are 90 degrees, 270 degrees.
Cosx=4/6
X=cos inverse of (4/6)= 36.869 degrees Cos is positive in 1st quadrant and 4th quadrant 360-36.869=323.131 Answers are 36.869 degrees and 323.131 degrees
This comment has been removed by the author.
ReplyDeleteSolving Quadratic Trigonometric Equations
ReplyDeleteQuadratic trigonometric equations are ones that involve products of trigonometric functions.
EXAMPLE 1: Solve the trigonometric equation 2cosx+2=3.
A). Solve for cos x – cos x = ½
B). Solve for x by finding all the values in the interval 0 < x < 360 that satisfy the equation. In
this case, with cosine being positive and equal to ½ , there are two values. One in the first
quadrant of the unit circle – x = pi/3.
C). Therefore, the second one is in the forth quadrant since cosine is positive there too. Making pi/3 negative and adding 360, your second value is 5pi/3.
D). Your answers are 60 degrees and 300 degrees.
EXAMPLE 2: Solve tan squared x-3=0.
A). Add 3 to the right side of the equal sign.
B). Since tan is squared, find the square root of tan squared and 3. – tan x = +/- square root of 3
C). Your answers are 60 degrees and 120 degrees.
*When dealing with a trig function that has a power, it will be solved by extracting square roots or factoring.
EXAMPLE 3: Solve 2sin squared x + sin x -1 = 0.
A). Since we’re dealing with just sin, you can factor.
B). (2sinx-1)(sinx+1) = 0
C). 2sinx-1 = 0 / sinx+1 = 0
D). sin x = ½ / sin x = -1
E). x = 30 degrees, 150 degrees and x = 270 degrees
**Note that many of the techniques used in solving algebraic equations are also used to solve trigonometric equations.
http://www.regentsprep.org/Regents/math/algtrig/ATT10/trigequations2.htm
http://www.analyzemath.com/Tutorial-Trigonometric-Equations/Tutorial-Trigonometric-Eq.html
This comment has been removed by the author.
ReplyDeleteThere are a few different ways that you can solve a quadratic formula. This is something that we learned in algebra 2, but it takes a bit of practice. Sometimes factoring can be used to solve a quadratic, but the quadratic formula will ALWAYS work every time. The formula is x = - b +/- square root (b^2- 4ac)/ 2a
ReplyDeleteExample 1: Solve x^2 + 3x – 4 = 0
a. Ths problem could be solved by factoring, but using the quadratic formula, you just plug it in. x = -3 +/- square root( (3) ^2 -4(1)(-4)/ 2(1)
b. Now, that it is plugged in, you just solve. -3 +/- square root(9 + 16)/ 2
c. -3 +/- square root(25)/2
d. -3 +/- 5/2
e. -3-5/2 and -3 + 5/ 2 ( Simply make two equations. )
f. -8/2 and 2/2 = -4, 1
x= -4 and x= 1
Example 2: x(x-2) =4
In order to use the quadratic formula, you must have everything equal to zero, and that is not the case in this problem.
a. x(x-2)= 4
b. x^2 -2x = 4
c. x^2 -2x -4 = 0
Now you can plug it into the formula.
a. x = -(-2) +/- square root( (-2) ^2 – 4(1)(-4) ) / 2(1)
b. 2 +/- square root(4 +16)/2
c. 2 +/- square root(20)/ 2
d. 2 +/- square root(4) square root(5)/ 2
e. 2 +/- 2 square root(5)/2
f. 2(1 +/- square root(5)/ 2(1) = 1 +/- square root(5)
g. You will get decimal answers, but you round them two decimal places.
x= -1.24 and x= 3.24
http://www.purplemath.com/modules/quadform2.htm
A trigonometric equation is one that involves one or more of the six functions sine, cosine, tangent, cotangent, secant, and cosecant. Some trigonometric equations, like x = cos x, can be solved only numerically. But a great many can be solved in closed form, and this page shows you how to do it in five steps.
ReplyDeleteContents:
Overview
Step 1. Get one function of one angle.
Step 2. Solve for the value(s) of a trig function.
Step 3. Solve for the angle.
Step 4. Solve for the variable.
Step 5. Apply any restrictions.
More Examples
What’s New
Copying: You’re welcome to print copies of this page for your own use, and to link from your own Web pages to this page. But please don’t make any electronic copies and publish them on your Web page or elsewhere.
Overview
If a trig equation can be solved analytically, these steps will do it:
Put the equation in terms of one function of one angle.
Write the equation as one trig function of an angle equals a constant.
Write down the possible value(s) for the angle.
If necessary, solve for the variable.
Apply any restrictions on the solution.
http://oakroadsystems.com/math/trigsol.htm
When the trig function has a power, it will have to be solved by extracting square roots or factoring.
Example 1:
Solution:
Example 2:
Solution:
At this point we know that :
Now, implies that (see graph). Since the sine function has maximum and minimum values of +1 and -1, has no solutions.
Thus the answer is the only solution.
http://www.regentsprep.org/Regents/math/algtrig/ATT10/trigequations2.htm
There are many ways to solve quadratic trig equations. The two main ways are completing the square and using the quadratic formula. I’m going to show you how to use the quadratic formula because we like to use formulas. The quadratic formula is something we learned in Algebra 2. The formula: x = -b +/- square root (b2 – 4ac)/2a
ReplyDeleteExample 1:
Solve
x2 + 2x –8 = 0
1. Plug into the formula. -2 +/- square root 22 - 4 (1) (-8) / 2(1)
2. Solve the equation. -2 +/- square root (4 + 32) / 2
3. -5 +/- square root (36) / 2
4. -5 +/- 2/2
5. -5 + 2/2, -5 – 2/2
6. -5 + 1 = -4, -5 – 1 = -6
Example 2:
Solve
x2 + 4x – 12 = 0
1. Plug into formula. -3 +/- square root 42 – 4 (1) (12) / 2(1)
2. Solve the equation. -3 +/- square root (16 + 48)
3. -3 +/- square root (64) / 2
4. -3 +/ - 8/2
5. -3 + 8/2, -3 – 8/2
6. -3 + 4 = 1, -3 – 4 = -7
http://mathforum.org/library/drmath/view/62884.html
There are a few different ways that you can solve a quadratic formula. We learned dealing with quadratic formula in algebra 2. Quadratic can be factored to solve. Quadratic formula will never fail for u. It will work every time.
ReplyDeleteFormula is the formula is x = - b +/- square root (b^2- 4ac)/ 2a.
Examples of Quadratic Formula:
Solve tan squared x-3=0.
A). Add 3 to the right side of the equal sign.
B). tan is squared; find the square root of tan squared and 3. – Tan x = +/- square root of 3
C). Your answers are 60 degrees and 120 degrees.
a = 1, b = –4, and c = –8.
X= -(-4) +/- square root of -4squared-4(1)(-8)/2(1)
4+/- square root 16+32/2= 4+/- square root 48/2
4+/- 4 square root of 3
2(2 +/- 2 square root 3/2 =2+/- 2 square root of 3
Which 2+/- 2 square roots of 3 is your answer.
http://www.purplemath.com/modules/solvquad4.htm
http://www.purplemath.com/modules/quadform2.htm
A trigonometric equation is one that involves one or more of the six functions sine, cosine, tangent, cotangent, secant, and cosecant. Some trigonometric equations, like x = cos x, can be solved only numerically. But a great many can be solved in closed form, and this page shows you how to do it in five steps.
ReplyDeleteEXAMPLE A:
3tan²(B/2) − 1 = 0
1.To solve it, add 1 to both sides and divide by 3: tan²(B/2) = 1/3
2.Then take square root of both sides:
Final answers: tan(B/2) = ±√(1/3) or ±√(3)/3
Be sure to remember to use the plus-or-minus sign ± when taking the square root of both sides; if you do not there may be some solutions you have overlooked because of it.
EXAMPLE B:
2 cos x + 2 = 3
1.solve for cos(x)
cos x = 1/2
2.solve for x by finding all values in the interval [0 , 2pi) that statisfy the above trigonometric equation.
3.In this case, with cosine positive and equal to 1 / 2, there are two values: one in the first quadrant of the unit circle.
x1 = pi / 3
and a second one in the fourth quadrant (see the two solutions in unit circle in figure below).
SOLUTIONS: x2 = 2*pi - pi / 3 = 5*pi / 3
http://www.analyzemath.com/Tutorial-Trigonometric-Equations/Tutorial-Trigonometric-Eq.html
http://oakroadsystems.com/math/trigsol.htm#Step2
I actually remember something about quadratic equations from algebra 2! There are 2 main ways, the quadratic formula and completing a square. Both are fairly easy.
ReplyDeleteFORMULA:
-b +/- square root b^2-4ac/2a
EXAMPLE:
x^2+2x-3=0
x=-2+/- square root of 2^2 -(4)(1)(-8)/2(1)
this is very simple because now you just solve the equation.
-2+/- (4)+(12)/2
-2+/- square root of 16/2
-2+4/2 and -2-4/2
when solving for a quadratic you must have 2 equations.
-2+2 and -2-2
x=0 and x=-4
VERY SIMPLE!
since you have a formula you easily just plug it in.
As long as you know this quadratic formula you're fine.
Trigonometry is ridiculous. There are so many ways to do the same thing. This is just another way to do just that.
ReplyDeleteSolve sin2(x) – sin(x) = 2 on 0° < x < 360°
This is a quadratic in sine, so I can apply some of the same methods:
sin2(x) – sin(x) – 2 = 0
(sin(x) – 2)(sin(x) + 1) = 0
sin(x) = 2 (not possible!) or sin(x) = –1
Only one of the factor solutions is sensible. For sin(x) = –1, I get:
x = 270°
Solve cos2(x) + cos(x) = sin2(x) on 0° < x < 360°
I can use a trig identity to get a quadratic in cosine:
cos2(x) + cos(x) = sin2(x)
cos2(x) + cos(x) = 1 – cos2(x)
2cos2(x) + cos(x) – 1 = 0
(2cos(x) – 1)(cos(x) + 1) = 0
cos(x) = 1/2 or cos(x) = –1
The first trig equation, cos(x) = 1/2, gives me x = 60° and x = 300°. The second equation gives me x = 180°. So my complete solution is:
x = 60°, 180°, 300°
http://www.purplemath.com/modules/solvtrig.htm
There isn't one way to solve a quadratic formula but the main way to solve it is the quadratic formula. Another way to do it would be to factor in the simplest situation ax^2 + bx + c and then factor. You can also use the complete the square method. But the way that it will always work is to use the quadratic formula. The quadratic formula says that x= -b +/- the square root of (b^2-4ac)/2a. You plug the numbers into the formula and come up with the answer. The key is to know the formula, which is also the key to Chapter 10(:.
ReplyDeleteExample 1: 4x^2 + 4x -2 = 0
a. Plug the correct numbers into the quadratic formula.
x= -4 +/- the square root of (4^2- 4(4)(-2)/2(4)
b.Multiply everything you can.
x= -4 +/- the square root of (16+32)
c. complete the problem
x= -4 +/- the square root of (48)/2
d. There are 2 answers a negative and a positive.
Example 2:
a. 3x^2+5x-2=0
b. x=-5 +/- the square root of (5^2-4(3)(-2)/2
c. 5 +/-the square root of (25+24)/2
d.5 +/- the square root (49)/2
e.(5+7)/2= 12/2=6
f. (5-7)/2= (-2/2)= -1
g. there are two answers in this problem because it is + or - x=6 and -1
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut17_quad.htm
In section ten four we did solved quadratic trig equations . The good thing about solving them is that you solve them the exact way you would solve a regular quadratic equation. Although the quadratic formula method may be used , the factoring method is more simple.
ReplyDeleteHere are some examples :
3sin^2x+sinx=-2
In order to solve a quadratic you must have the equation equal to zero. In order to do so you must bring the negative two over to get: 3sin^2x+sinx+2=0 Now you would solve as if you would a regular quadratic equation.
Here is an example of solving regular quadratic equations. You must first know how to solve a regular quadratic equation before you will be able to solve one with a trig function.
Example number one : very simple..
3x^2 + 12x=-9 first bring th negative nine over..
3x^2+12x+9=0 next step is to factor.. factors of 3(9) that add to = 12; (9,3)
3x^2+3x+9x+9=0 factor out..
3x(x+1)+9(x+1) now u solve.. set = to 0
3x+9=0 x+1=0
X=-3 x=-1
Example number two:
3sin^2x+sinx+2=0
=3 sin^2x +3sinx+2sinx+2=0 the next step is to factor..
=3sinx(sinx+1)2(sinx+1) set both equations equal to zero..
(3sinx+2)=0 (sinx+1)=0 now you will solve..
Sinx=3/2 sin x=1
X=sin-1(3/2) x=sin-1(1)
Find theinverses and you have your answers..
Simple enough :)
When solving a quadratic for a trig equation you go back to the beginning ways of solving a quadratic, which is what you learn in Algebra 2. You can either:
ReplyDelete1.Complete the square
Solve x2 - 6x + 2 = 0 by completing the square
x2 - 6x = -2
[To complete the square on the LHS (left hand side), we must add 62/4 = 9. We must, of course, do this to the RHS also].
∴ x2 - 6x + 9 = 7
∴ (x - 3)2 = 7
[Now take the square root of each side]
∴ x - 3 = ±2.646 (the square root of 7 is +2.646 or -2.646)
∴ x = 5.646 or 0.354
2.Use the quadratic formula
ax2 + bx + c = 0 ---> x= -b± √( b2-4ac)/2a
3.Factor
Solve x2 + 2x - 8 = 0
∴ (x - 2)(x + 4) = 0
∴ either x - 2 = 0 or x + 4 = 0
∴ x = 2 or x = - 4
When given in the form of a trig equation you do the same..
1. Some of these are copied from each other. You will not get credit if you have copied any part of some else's blog.
ReplyDelete2. Many of you forgot there are two angles for each inverse with exceptions at 1 and -1. sin^-1 (1/2) has two quadrants where sin is positive.
3. If you gave examples without trig functions redo them with trig functions in a new comment to get credit.
4. We NEVER used the quadratic formula to solve a trig equation. You need to revisit your notes on these. This is why you are having trouble with 8-4 & 10-4. Relook at the example problems from those sections.
This week we learned how solve quadratic trig equation is section 10-4. Section 10-4 is very similar to section 8-4 but the only difference is that 10-4 involves all 8-4 formulas when you need to plug in to replace and all the formulas in all of chapter 10. Remember there are six functions in a trigonometry equation.
ReplyDeleteThe six functions are:
Sine
Cosine
Tanganent
Cotangent
Cosecant
Secant
Examples:
Cos2x=1-sinx for 0< equal to x< equal to 360 degrees
-2sinx^2+sinx=0
Sinx(-2sinx+1)=0
Sinx=0
X=sin inverse of 0
Equal to 0,180 degrees, 360 degrees=answer also
Sinx=1/2
X=sin invere(1/2)
Equal to 30 degrees where sine is +
1st quadrant and 2nd quadrant
180-30=150
Answers are 150 degrees, 30 degrees
3cos2x+cosx=2 for 0<equal to x<equal to 360 degrees
3(cos^2x-1)+cosx+/-2
6cos^2x-3+cosx=2
6 cos^2x+ cosx-5=0
Cosx(6cosx+1)-5=0
Cosx(6cosx+1)=5
Cosx=0
X=cos inverse of 0 =90 degrees
Answers are 90 degrees, 270 degrees.
Cosx=4/6
X=cos inverse of (4/6)= 36.869 degrees
Cos is positive in 1st quadrant and 4th quadrant
360-36.869=323.131
Answers are 36.869 degrees and 323.131 degrees