JEE Main 2023 — Mathematics Trigonometry

Given,

$$3{\mathrm{cos}}^{4}\theta -5{\mathrm{cos}}^{2}\theta -2{\mathrm{sin}}^{6}\theta +2=0$$3cos4θ−5cos2θ−2sin6θ+2=0

$$\Rightarrow {\mathrm{cos}}^{2}\theta [3{\mathrm{cos}}^{2}\theta -5]-2{\mathrm{sin}}^{6}\theta +2=0$$⇒cos2θ[3cos2θ−5]−2sin6θ+2=0

$$\Rightarrow (1-{\mathrm{sin}}^{2}\theta )(3-3{\mathrm{sin}}^{2}\theta -5)-2{\mathrm{sin}}^{6}\theta +2=0$$⇒(1−sin2θ)(3−3sin2θ−5)−2sin6θ+2=0

$$\Rightarrow ({\mathrm{sin}}^{2}\theta -1)(3{\mathrm{sin}}^{2}\theta +2)-2{\mathrm{sin}}^{6}\theta +2=0$$⇒(sin2θ−1)(3sin2θ+2)−2sin6θ+2=0

Let ${\mathrm{sin}}^{2}\theta =t$sin2θ=t

$$\Rightarrow (t-1)(3t+2)-2{t}^{3}+2=0$$⇒(t−1)(3t+2)−2t3+2=0

$$\Rightarrow (t-1)(3t+2)-2({t}^{3}-1)=0$$⇒(t−1)(3t+2)−2(t3−1)=0

We know that ${a}^{3}-{b}^{3}=(a-b)({a}^{2}+ab+{b}^{2})$a3−b3=(a−b)(a2+ab+b2)

$$\Rightarrow (t-1)(3t+2)-2(t-1)({t}^{2}+t+1)=0$$⇒(t−1)(3t+2)−2(t−1)(t2+t+1)=0

$$\Rightarrow (t-1)[3t+2-2({t}^{2}+t+1)]=0$$⇒(t−1)[3t+2−2(t2+t+1)]=0

$$(t-1)[2{t}^{2}-t]=0$$(t−1)[2t2−t]=0

$$\Rightarrow t=0,1,\frac{1}{2}$$⇒t=0,1,21​

For ${\mathrm{sin}}^{2}\theta =0$sin2θ=0, $\theta =0,\pi ,2\pi$θ=0,π,2π

${\mathrm{sin}}^{2}\theta =1$sin2θ=1, $\theta =\frac{\pi }{2},\frac{3\pi }{2}$θ=2π​,23π​

${\mathrm{sin}}^{2}\theta =\frac{1}{2}$sin2θ=21​, $\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$θ=6π​,65π​,67π​,611π​

$\therefore$∴ Total solution $=9$=9

Hence this is the required option.