1. ## Trig Identities

Solve for $0\leq x \leq 2\pi$

well i am bad at the math tool, but its 0<=x<=2pi

Cos2x = 1 - sinx

What would be the best way to set this up?

Solve for $0\leq x \leq 2\pi$

well i am bad at the math tool, but its 0<=x<=2pi

Cos2x = 1 - sinx

What would be the best way to set this up?

Use the identity that $\cos(2x) = 1 - 2\sin^2x$.

Get:

$1-2\sin^2x = 1 - \sin x$.

So $2 \sin^2 x - \sin x = 0$.

Factor:
$\sin x \cdot (2 \sin x - 1) = 0$.

Can you finish it from here?

3. Yes, I got to x=30 which I converted to radians...so $\pi/6$

4. Yes $\pi/6$ is a solution.
There are several answers in that interval given $(0 \le x \le 2\pi)$.

Note that $5\pi/6$ is also an answer, as are $0, \pi$, and $2\pi$.

Do you see why?