solve for x where 0 ≤ x ≤ 2π. cos²x = 2 sin x + 2

Answer:

$x = \frac{3\pi}{2}$

Explanation:

Step1: Use trig identity

Recall $\cos^{2}x=1 - \sin^{2}x$. So the equation becomes $1-\sin^{2}x=2\sin x + 2$.

Step2: Rearrange to quadratic form

Rearrange to get $\sin^{2}x+2\sin x + 1 = 0$. Let $t=\sin x$, then $t^{2}+2t + 1=0$.

Step3: Factor quadratic

Factor $t^{2}+2t + 1$ as $(t + 1)^{2}=0$. So $t=\sin x=-1$.

Step4: Solve for x in given range

Since $0\leq x\leq2\pi$ and $\sin x=-1$, then $x=\frac{3\pi}{2}$.