1. ## Solving for theta

Solve sin^2ϴ = 1 + cosϴ 0 ≤ ϴ ≤ 2pi

I dont even know where to start! any help would be appreciated thank you

2. Originally Posted by cordelljheckel
Solve sin^2ϴ = 1 + cosϴ 0 ≤ ϴ ≤ 2pi

I dont even know where to start! any help would be appreciated thank you
$sin^2 \theta = 1-cos^2 \theta$

Sub in, 1 will cancel and solve the quadratic

3. so am i substituting sin^2(x) with 1 - cos2x / 2 then solving?

4. No, that wasn't the given indication. Read it again.

5. i dont understand, which identity are you using to sub in?

6. Originally Posted by cordelljheckel
i dont understand, which identity are you using to sub in?
Just like e^(i*pi) said,

$sin^2(\theta) = 1 - cos^2(\theta)$

Substitute into the first equation to get:

$1-cos^2\theta = 1 + cos\theta$

Now solve.