Question

Please help me thank you

Answer 1

`\theta = 0 ,(7\pi)/(6) ,(11\pi)/(6 )`

Step-by-step explanation

We want to solve

`\sin( \theta) + 1 = \cos(2 \theta)`

on the interval

`0 \leqslant \theta \: < \: 2\pi`

Use the double angle identity to obtain:

`\sin( \theta) + 1 = 1 - 2\sin ^(2) \theta`

Simplify to get;

`2\sin ^(2) \theta + \sin( \theta) + 1 - 1 = 0`

`2\sin ^(2) \theta + \sin( \theta) = 0`

Factorize to obtain:

`\sin \theta (2\sin \theta + 1) = 0`

Either

`\sin \theta = 0`

This gives us

`\theta = 0`

on the given interval.

Or

`2\sin \theta + 1= 0`

`\sin \theta = - (1)/(2)`

This gives us

`\theta = (7\pi)/(6) ,(11\pi)/(6 )`

Therefore the solutions within the interval are:

`\theta = 0 ,(7\pi)/(6) ,(11\pi)/(6 )`