Mathematics - Trigonometric Equation Question with Solution | TestHub

${\sin }^{2}x-2\sin x+5=(\sin x-1{)}^2+4\ge 4$

$\therefore 2^{\sqrt{{\sin }^{2}x-2\sin x+5}}\ge 2^2=4$

and ${\sin }^{2}y\le 1\Rightarrow \frac{1}{4^{{\sin }^{2}y}}\ge \frac{1}{4}$

∴ LHS $\ge 1$ and according to question LHS $\le 1$,

so therefore, LHS = 1 for which

$\begin{array}{rl}& \therefore {\sin }^{2}x-2\sin x+5=4\\ & (\sin x-1{)}^2=0\\ & \sin x=1\Rightarrow x=(2n+1)\frac{\pi }{2}\end{array}$

and ${\mathrm{cosec}}^{2}y=1,{\sin }^{2}y=1$ or $\cos y=0$

$y=(2m+1)\frac{\pi }{2}$