JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Trigonometric Equations

Since ${\sin }^{4}x\ge 0$, the equation $1+{\sin }^{4}x={\cos }^{2}3x$ forces ${\cos }^{2}3x\ge 1$. But a cosine-squared can never exceed $1$, so it must equal exactly $1$, which in turn forces ${\sin }^{4}x=0$, i.e. $\sin x=0$. Then $x=n\pi$, and these automatically make $3x=3n\pi$ a multiple of $\pi$ so ${\cos }^{2}3x=1$ holds too. Within $\left[-\frac{5\pi }{2},\frac{5\pi }{2}\right]$ the multiples of $\pi$ are $x=-2\pi ,-\pi ,0,\pi ,2\pi$, which is $5$ solutions.