JEE Main 2021 — Mathematics Trigonometry

We know that $0\leq {\mathrm{sin}}^{2}x\leq 1$0≤sin2x≤1 and $0\leq {\mathrm{cos}}^{2}x\leq 1$0≤cos2x≤1

Also, we know for a number which lie between $0$0 to $1,$1, higher the power of the number the less is its value.

$$\Rightarrow {\mathrm{sin}}^{7}x\leq {\mathrm{sin}}^{2}x\leq 1\ldots (1)$$⇒sin7x≤sin2x≤1…(1)

and ${\mathrm{cos}}^{7}x\leq {\mathrm{cos}}^{2}x\leq 1\ldots (2)$cos7x≤cos2x≤1…(2)

Also, we know that ${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$sin2x+cos2x=1

This means the equality must hold for $(1)$(1) and $(2)$(2)

$\Rightarrow {\mathrm{sin}}^{7}x={\mathrm{sin}}^{2}x$⇒sin7x=sin2x and ${\mathrm{cos}}^{7}x={\mathrm{cos}}^{2}x$cos7x=cos2x

$\Rightarrow {\mathrm{sin}}^{2}x({\mathrm{sin}}^{5}x-1)=0$⇒sin2x(sin5x−1)=0 and ${\mathrm{cos}}^{2}x({\mathrm{cos}}^{5}x-1)=0$cos2x(cos5x−1)=0

$\Rightarrow \mathrm{sin}x=0$⇒sinx=0 or $\mathrm{sin}x=1$sinx=1 and $\mathrm{cos}x=0$cosx=0 or $\mathrm{cos}x=1$cosx=1$\Rightarrow x=0,2\pi ,4\pi ,\frac{\pi }{2},\frac{5\pi }{2}.$⇒x=0,2π,4π,2π​,25π​.

Thus, there are total $5$5 solutions.