Expert: Paul Klarreich - 1/1/2007

Question
SOLVE for X in each of the following:
a) X is between 0 and 2pi (meaning show answer in radians??)
2sin^2 x = 2 + cos2x

I substituted using a double angle formula: cos2A=1-2sin^2 A.
cos 2x=2sin^2 x -2
1-2sin^2 x=2sinx-2
-2sin^2 x=2sinx-3
2sin^2 x +2sinx-3=0
Let y= sinx
2y^2 + 2y-3=0
I was unable to factor this and I used the quadratic formula. I ended up with x= -1 plus/minus square root of 7 divided by 2. is this anywhere close? If so what do I do next?


b) X is between 0 degrees and 360 degrees (meaning show answer in degrees??)
sin2x + cosx = 0

sin2x=-cosx
2sinx cosx= -cosx
2sinx + cos2x=0
2sinx + (1-2 sin^2 x) =0
2sinx + 1-2 sin^2 x =0
----I divided by two on both sides-----
sinx - sin^2 x = 0

Did I do anything wrong? If I did it right then what do I do next?


c) X is between 0 degrees and 360 degrees (meaning show answer in degrees??)
3sin(x + 20) degrees = -1

3sin(x + 20)= -1
---I divided both sides by 3---
sin(x+20)= -1/3
Now what? I think that I need to find a reference angle or something like that...

Please help me!!

I understand that this is quite a lenghty question but any help you can offer would be GREATLY appreciated. Thank you in advance for your time.  

Answer
Questioner:  Jon
Category:  Advanced Math
 
Subject:  Pre Calculus Trig. Solving Equations
Question:  SOLVE for X in each of the following:
a) X is between 0 and 2pi (meaning show answer in radians??)
2sin^2 x = 2 + cos2x

I substituted using a double angle formula: cos2A=1-2sin^2 A.
cos 2x=2sin^2 x -2
1-2sin^2 x=2sinx-2
-2sin^2 x=2sinx-3
2sin^2 x +2sinx-3=0
Let y= sinx
2y^2 + 2y-3=0
I was unable to factor this and I used the quadratic formula. I ended up with x=

-1 plus/minus square root of 7 divided by 2. is this anywhere close? If so what do I do next?


b) X is between 0 degrees and 360 degrees (meaning show answer in degrees??)
sin2x + cosx = 0

sin2x=-cosx
2sinx cosx= -cosx
2sinx + cos2x=0  <<< doesn't look right.
2sinx + (1-2 sin^2 x) =0
2sinx + 1-2 sin^2 x =0
----I divided by two on both sides-----
sinx - sin^2 x = 0

Did I do anything wrong? If I did it right then what do I do next?


c) X is between 0 degrees and 360 degrees (meaning show answer in degrees??)
3sin(x + 20) degrees = -1

3sin(x + 20)= -1
---I divided both sides by 3---
sin(x+20)= -1/3
Now what? I think that I need to find a reference angle or something like

that...

Please help me!!

I understand that this is quite a lenghty question but any help you can offer

would be GREATLY appreciated. Thank you in advance for your time.
........................................................
A. I think you messed up the algebra here.  It goes like this:

(ORIG) 2 sin^2 x = 2 + cos2x

FORMULA:  cos 2x = 1 - 2 sin^2 x
Substitute: (writing S for sin x)
2S^2 = 2 + 1 - 2S^2
4S^2 = 3
sin x = sqrt(3)/2
Now you should be OK.  You will have things like pi/3, etc.

B. sin2x + cosx = 0
Try:  sin 2x = 2 sin x cos x

2 sin x cos x + cos x = 0
Factor. (Yes, you can factor even if you have more than one function sitting

there.)

cos x (2 sin x + 1) = 0
Set each factor equal to zero;

cos x = 0, which will give you 90 and 270.

2 sin x + 1 = 0, which will become sin x = -1/2.  Now your reference angle is 30

and you have solutions in the 3rd and 4th quadrants.

C. 3sin(x + 20)= -1

  sin(x+20)= -1/3

So far, so good.  Now you will have to find a reference angle.  Like this:

1/3 = 0.3333..

x+20 = INV SIN (0.3333) = 19.5 (degrees)

Angle in Q3:  189.5 = x + 20,  x = 169.5
Angle in Q4:  340.5 = x + 20,  x = 320.5