Question

Help this is for calc

Answer 1

theta=pi/4, 3pi/4, 5pi/4, 7pi/4

Explanation:

Answer 2

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

`2\sin^2\theta=1\implies\sin^2\theta=\frac12\implies\sin\theta=\pm\frac1{\sqrt2}`

Then

`\theta=\frac\pi4+2n\pi,\frac{3\pi}4+2n\pi,-\frac\pi4+2n\pi,\text{ or }-\frac{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=\frac\pi4,\frac{3\pi}4,\frac{5\pi}4,\frac{7\pi}4}`

Or, you can use the double angle identity:

`\sin^2\theta=\frac{1-\cos(2\theta)}2=\frac12\implies1-\cos(2\theta)=1\implies\cos(2\theta)=0`

Then

`2\theta=\frac\pi2+n\pi\implies\theta=\frac\pi4+\frac{n\pi}2`

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