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Q.
The number of solutions of the equation $cot^{2} sin x + 3=1$ in $0,3 \pi $ is equal to
NTA AbhyasNTA Abhyas 2022
Solution:
$cot^{2} sin x + 3=1=cot^{2} \frac{\pi }{4}$
$\Rightarrow sin x+3=n\pi \pm\frac{\pi }{4}$
Also, $2\leq sin x+3\leq 4$
$\Rightarrow sin x+3=\pi -\frac{\pi }{4},\pi +\frac{\pi }{4}$
$\Rightarrow sin x=\frac{3 \pi }{4}-3$ or $\frac{5 \pi }{4}-3$
