If cos^2 x + sin x + 1 = 0, and 0 < x < 2π, then x =

Question

\[ \text{If } \cos^2x+\sin x+1=0, \]

\[ 0<x<2\pi, \]

\[ \text{then } x= \]

Using identity

\[ \cos^2x=1-\sin^2x \]

Substituting,

\[ 1-\sin^2x+\sin x+1=0 \]

\[ -\sin^2x+\sin x+2=0 \]

\[ \sin^2x-\sin x-2=0 \]

\[ (\sin x-2)(\sin x+1)=0 \]

Since

\[ \sin x\neq2 \]

therefore,

\[ \sin x=-1 \]

Now,

\[ x=\frac{3\pi}{2} \]

\[ \boxed{\frac{3\pi}{2}} \]

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