# Math Help - [SOLVED] Sin, cos! Urgent!

1. ## [SOLVED] Sin, cos! Urgent!

Barron Page 77

Solve for 2sinx+ cos2x=2sin^2x-1 for 0< and equals x < and equals 2 $\pi$?

Thanks in advance! Urgent!!

2. Hint: $\cos 2x = 1 - 2\sin^2 x$

Arrange your equation and you will have it in the form of: $ay^2 + by + c = 0$

Then you can use the quadratic formula.

3. Ok thanks.

I got 4sin^2x-sinx+2=0
do i just solve x? I don't have to consider sin?

4. You solve the quadratic taking sin(x) as the variable and then you find x with
x = arcsin(sin(x)) there can be 2 solutions.

5. How do you when to use arcsin or when to use sin?

6. Suppose
$a\cdot sin(x)^2 + b \cdot sin(x) + c = 0$
$sin(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x = arcsin(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a})$