Expert: Paul Klarreich - 11/15/2010

Question
How do I solve these math problems:

cos^2x = sinx, when x <0, 2pi>

and

4sin((pi/12)x)+4cos((pi/12)x) = 0, when x <0, 24>

Answer
Questioner:   Anette  
Country:  Norway
Category:  Advanced Math
Private:  No
 
Subject:  triogometry
Question:  How do I solve these math problems:

cos^2x = sinx, when x <0, 2pi>

and

4sin((pi/12)x)+4cos((pi/12)x) = 0, when x <0, 24>
....................................................
For this one,

cos^2x = sinx  

use an identity to write:

1 - sin^2x = sinx

then
0 = - 1 + sin^2x + sinx

0 =  sin^2x + sinx - 1
       - 1 +-sqrt(5)
sin x = --------------
               2

Now you will need a calculator; you should get 4 solutions.
...........................
4sin((pi/12)x)+4cos((pi/12)x) = 0

sin((pi/12)x) + cos((pi/12)x) = 0

sin((pi/12)x) = - cos((pi/12)x)

Divide and use another identity:

tan((pi/12)x) = -1

That will give you

(pi/12)x = 3pi/4  and 7pi/4

You can finish up.