ANSWER
[tex]\theta = 0 ,\frac{7\pi}{6} ,\frac{11\pi}{6 } [/tex]
EXPLANATION
We want to solve
[tex] \sin( \theta) + 1 = \cos(2 \theta) [/tex]
on the interval
[tex]0 \leqslant \theta \: < \: 2\pi[/tex]
Use the double angle identity to obtain:
[tex] \sin( \theta) + 1 = 1 - 2\sin ^{2} \theta[/tex]
Simplify to get;
[tex] 2\sin ^{2} \theta + \sin( \theta) + 1 - 1 = 0[/tex]
[tex]2\sin ^{2} \theta + \sin( \theta) = 0[/tex]
Factorize to obtain:
[tex]\sin \theta (2\sin \theta + 1) = 0[/tex]
Either
[tex]\sin \theta = 0[/tex]
This gives us
[tex] \theta = 0[/tex]
on the given interval.
Or
[tex]2\sin \theta + 1= 0[/tex]
[tex]\sin \theta = - \frac{1}{2} [/tex]
This gives us
[tex]\theta = \frac{7\pi}{6} ,\frac{11\pi}{6 } [/tex]
Therefore the solutions within the interval are:
[tex]\theta = 0 ,\frac{7\pi}{6} ,\frac{11\pi}{6 } [/tex]
