JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Trigonometric Equations

We work with the equation $3{\cos }^4\theta -5{\cos }^2\theta -2{\sin }^6\theta +2=0$ on $[0,2\pi ]$. Split the middle term as $-5{\cos }^2\theta =-3{\cos }^2\theta -2{\cos }^2\theta$ and use $2-2{\cos }^2\theta =2{\sin }^2\theta$, so the equation becomes $3{\cos }^2\theta ({\cos }^2\theta -1)+2{\sin }^2\theta (1-{\sin }^4\theta )=0$. Since ${\cos }^2\theta -1=-{\sin }^2\theta$ and $1-{\sin }^4\theta ={\cos }^2\theta (1+{\sin }^2\theta )$, this factorises neatly as ${\sin }^2\theta {\cos }^2\theta (2{\sin }^2\theta -1)=0$. Hence ${\sin }^2\theta =0$ gives $\theta =0,\pi ,2\pi$, which are three values; ${\cos }^2\theta =0$ gives $\theta =\frac{\pi }{2},\frac{3\pi }{2}$, which are two values; and $2{\sin }^2\theta =1$ gives $\theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$, which are four values. Collecting every admissible root in $[0,2\pi ]$, the set $\mathrm{S}$ contains $9$ elements.