JEE Main 2023 — Trigonometric Equations Question with Solution

Given,$f\left(\theta \right)=3\left({\sin }^4\left(\frac{3\pi }{2}-\theta \right)+{\sin }^4(3\mathrm{\pi }+\theta )\right)-2\left(1-{\sin }^{2}2\theta \right)$

and $S=\left\{\theta \in [0,\pi ]:f^{\text{'}}\left(\theta \right)=-\frac{\sqrt{3}}{2}\right\}$

$f\left(\theta \right)=3\left({\sin }^4\left(\frac{3\pi }{2}-\theta \right)+{\sin }^4(3\mathrm{\pi }+\theta )\right)-2\left(1-{\sin }^{2}2\theta \right)$

$\Rightarrow f\left(\theta \right)=3\left({\cos }^4\theta +{\sin }^4\theta \right)-2 {\cos }^{2}2\theta$

$\Rightarrow \mathrm{f}\left(\theta \right)=3\left(1-\frac{1}{2}{\sin }^{2}2\theta \right)-2 {\cos }^{2}2\theta$

$\Rightarrow \mathrm{f}\left(\theta \right)=3-\frac{3}{2}{\sin }^{2}2\theta -2 {\cos }^2\theta$

$\Rightarrow f\left(\theta \right)=\frac{3}{2}-\frac{1}{2}{\cos }^{2}2\theta =\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4\theta }{2}\right)$

$\Rightarrow f\left(\theta \right)=\frac{5}{4}-\frac{\cos 4\theta }{4}$

Now on differentiating we get, $f^{\text{'}}\left(\theta \right)=\sin 4\theta$

$\Rightarrow f^{\text{'}}\left(\theta \right)=\sin 4\theta =-\frac{\sqrt{3}}{2}$

$\Rightarrow 4\theta =\mathrm{n}\pi +(-1{)}^{\mathrm{n}}\frac{\pi }{3}$

$\Rightarrow \theta =\frac{\mathrm{n}\pi }{4}+(-1{)}^{\mathrm{n}}\frac{\pi }{12}$

$\Rightarrow \theta =\frac{\pi }{12},\left(\frac{\pi }{4}-\frac{\pi }{12}\right),\left(\frac{\pi }{2}+\frac{\pi }{12}\right),\left(\frac{3\pi }{4}-\frac{\pi }{12}\right)$

$\Rightarrow 4\beta =\frac{\pi }{12}+\frac{\mathrm{\pi }}{6}+\frac{7\pi }{12}+\frac{8\pi }{12}=\frac{3\pi }{2}$

$\Rightarrow \beta =\frac{3\pi }{8}\Rightarrow f\left(\beta \right)=\frac{5}{4}-\frac{\cos \frac{3\pi }{2}}{4}=\frac{5}{4}$