SOLUTION
Write out the equation given
[tex]\cos ^2\theta=2+2\sin \theta[/tex]
Rearrange the equation, we have
[tex]\cos ^2\mleft(\theta\mright)-2-2\sin \mleft(\theta\mright)=0[/tex]
Then recall from trigonometry identies that
[tex]\cos ^2\theta=1-\sin ^2\theta[/tex]
Then substitute into the expression we have
[tex]\begin{gathered} 1-\sin ^2\theta-2-2\sin \theta=0 \\ \text{Then} \\ -\sin ^2\theta-2\sin \theta-1=0 \\ \text{Multiply through by -1, we have } \\ \sin ^2\theta+2\sin \theta+1=0 \end{gathered}[/tex]
Then
Replace
[tex]\begin{gathered} \sin ^2\theta=p,\text{ we have } \\ p^2+2p+1=0 \end{gathered}[/tex]
Then, solve by factorization, we have
[tex]\begin{gathered} p^2+p+p+1=0 \\ p(p+1)+1(p+1)=0 \\ (p+1)(p+1)=0 \end{gathered}[/tex]
Then we have that
[tex]\begin{gathered} (p+1)^2=0 \\ \text{Then } \\ p+1=0 \\ p=-1 \end{gathered}[/tex]
hence, we have that
[tex]\sin \theta=-1[/tex]
Therefore,
[tex]\theta=\sin ^{-1}(-1)[/tex]
Since the interval is
[tex]\begin{gathered} \lbrack0.2\pi) \\ \text{Then } \\ \theta=\frac{3\pi}{2}\text{ in radians } \end{gathered}[/tex]
Therefore
Answer: Θ = 3π/2 rads
