Question

Help this is for calc

Answer 1

You can solve for $\sin^2\theta$, then take square roots:

$2\sin^2\theta=1\implies\sin^2\theta=\dfrac12\implies\sin\theta=\pm\dfrac1{\sqrt2}$

Then

$\theta=\dfrac\pi4+2n\pi,\dfrac{3\pi}4+2n\pi,-\dfrac\pi4+2n\pi,\text{ or }-\dfrac{3\pi}4+2n\pi$

where $n$ is any integer; we get the following solutions in the interval $0\le\theta\le2\pi$:

$\boxed{\theta=\dfrac\pi4,\dfrac{3\pi}4,\dfrac{5\pi}4,\dfrac{7\pi}4}$

Or, you can use the double angle identity:

$\sin^2\theta=\dfrac{1-\cos(2\theta)}2=\dfrac12\implies1-\cos(2\theta)=1\implies\cos(2\theta)=0$

Then

$2\theta=\dfrac\pi2+n\pi\implies\theta=\dfrac\pi4+\dfrac{n\pi}2$

where $n$ is any integer, and we get the same solutions as above within the prescribed interval.