JEE Main 2026 Apr 2 Shift 2, Mathematics Q13: Trigonometric Equations

Factor the difference of fourth powers: ${\cos }^8\theta -{\sin }^8\theta =({\cos }^4\theta +{\sin }^4\theta )({\cos }^2\theta +{\sin }^2\theta )({\cos }^2\theta -{\sin }^2\theta )=({\cos }^4\theta +{\sin }^4\theta )\cos 2\theta$.

Multiplying by $2\sec 2\theta$ gives $2({\cos }^4\theta +{\sin }^4\theta )=2(1-2{\sin }^2\theta {\cos }^2\theta )=2-\frac{{\sin }^{2}2\theta }{1}\cdot \frac{1}{1}$. Cleanly: $2-{\sin }^{2}2\theta$.

So $a^2=2-{\sin }^{2}2\theta$. As ${\sin }^{2}2\theta \in [0,1]$, we have $a^2\in [1,2]$.

For integer $a$, $a^2=1$ requires ${\sin }^{2}2\theta =1$, i.e. $2\theta =(2n+1)\frac{\pi }{2}$, giving $\theta =(2n+1)\frac{\pi }{4}$. But these values make ${\tan }^2\theta =1$, which are excluded from $P$.

Hence no integer $a$ satisfies the condition: $n(S)=0$.