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Q.
The number of solution(s) of the trigonometric equation $sec^{2}x+cosec^{2}x-sin2x-3=0,$ in $\left[0 ,4 \pi \right]$ , is
NTA AbhyasNTA Abhyas 2022
Solution:
$sec^{2}x+cosec^{2}x=sin2x+3$
$1+tan^{2}x+1+cot^{2}x=1+sin2x+2$
$2+tan^{2}x+cot^{2}x=\left(sin x+cos x\right)^{2}+2$
Range of LHS is $\left[\right.4,\infty \left]\right.$
Range of RHS is $\left[\right.2,4\left]\right.$
So solution will exist when
$sin2x=1$ and $tanx=cotx=\pm1$
$2x=\frac{\pi }{2}\left(4 n + 1\right)$
$x=\frac{\pi }{4},\frac{5 \pi }{4},\frac{9 \pi }{4},\frac{13 \pi }{4}$
at these $x$ the value of $tanx=cotx=1$ so number of solution $=4$