If cos

Given,

$\cos ^{2} x+\sin x+1=0$ and $0

$1-\sin ^{2} x+\sin x+1=0$

i. e $(1-\sin x)(1+\sin x)+(1+\sin x)=0$

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

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

i. e $\sin x=-1$ or $\sin x=2$

since $\sin x \neq 2$

$\Rightarrow \sin x=-1 \quad$ (only possibility)

i. e $x=\frac{3 \pi}{2}$