(-cos²x)+sin x + 1 = 0

Explanation:

Step1: Use trigonometric identity

Recall $\cos^{2}x = 1-\sin^{2}x$. The equation $-\cos^{2}x+\sin x + 1=0$ becomes $-(1 - \sin^{2}x)+\sin x + 1 = 0$.

Step2: Simplify the equation

Expand the left - hand side: $-1+\sin^{2}x+\sin x + 1=0$. Then $\sin^{2}x+\sin x=0$.

Step3: Factor the equation

Factor out $\sin x$: $\sin x(\sin x + 1)=0$.

Step4: Solve for $x$

Set each factor equal to zero. If $\sin x=0$, then $x = k\pi$, where $k\in\mathbb{Z}$. If $\sin x+1 = 0$, i.e., $\sin x=-1$, then $x=\frac{3\pi}{2}+2k\pi$, $k\in\mathbb{Z}$.

Answer:

$x = k\pi$ or $x=\frac{3\pi}{2}+2k\pi,k\in\mathbb{Z}$