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A solution of the equation cos

Question:

A solution of the equation $\cos ^{2} x+\sin x+1=0$, lies in the interval

(a) $(-\pi / 4, \pi / 4)$

(b) $(\pi / 4,3 \pi / 4)$

(c) $(3 \pi / 4,5 \pi / 4)$

(d) $(5 \pi / 4,7 \pi / 4)$

Solution:

(d) $(5 \pi / 4,7 \pi / 4)$

Given:

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

$\Rightarrow\left(1-\sin ^{2} x\right)+\sin x+1=0$

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

$\Rightarrow \sin ^{2} x-\sin x-2=0$

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

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

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

$\Rightarrow \sin x-2=0$ or $\sin x+1=0$

$\Rightarrow \sin x=2$ or $\sin x=-1$

$\sin x=2$ is not possible.

$\Rightarrow \sin x=-1$

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

$\Rightarrow x=n \pi+(-1)^{n} \frac{3 \pi}{2}, \mathrm{n} \in \mathrm{Z}$

The values of x">x lies in the third and fourth quadrants.

Hence, $x$ lies in $\left(\frac{5 \pi}{4}, \frac{7 \pi}{4}\right)$.

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