Question:
If cos2x + sin x + 1 = 0, and 0 < x < 2π then x = _________.
Solution:
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 $x=\frac{3 \pi}{2}$
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