JEE Main 2020 — Mathematics Calculus

$$2{\mathrm{cot}}^{2}⁡\theta -\frac{5}{\mathrm{sin}⁡\theta }+4=0$$2cot2⁡θ−sin⁡θ5​+4=0

$$\frac{2{\mathrm{cos}}^{2}⁡\theta }{{\mathrm{sin}}^{2}⁡\theta }-\frac{5}{\mathrm{sin}⁡\theta }+4=0$$sin2⁡θ2cos2⁡θ​−sin⁡θ5​+4=0

$$2{\mathrm{cos}}^{2}⁡\theta -5\mathrm{sin}⁡\theta +4{\mathrm{sin}}^{2}⁡\theta =0,\mathrm{sin}⁡\theta \neq 0$$2cos2⁡θ−5sin⁡θ+4sin2⁡θ=0,sin⁡θ=0

$$2{\mathrm{sin}}^{2}⁡\theta -5\mathrm{sin}⁡\theta +2=0$$2sin2⁡θ−5sin⁡θ+2=0

$$(2\mathrm{sin}⁡\theta -1)(\mathrm{sin}⁡\theta -2)=0$$(2sin⁡θ−1)(sin⁡θ−2)=0

$$\mathrm{sin}⁡\theta =\frac{1}{2}$$sin⁡θ=21​

$\theta =\frac{\pi }{6},\frac{5\pi }{6};$θ=6π​,65π​; $\theta \in (0,2\pi )-{\pi }$θ∈(0,2π)−π

$$\therefore \int _{\pi /6}^{5\pi /6}{\mathrm{cos}}^{2}⁡3\theta d\theta =\int _{\pi /6}^{5\pi /6}\frac{1+\mathrm{cos}⁡6\theta }{2}d\theta$$∴∫π/65π/6​cos2⁡3θdθ=∫π/65π/6​21+cos⁡6θ​dθ

$$=\frac{1}{2}{[\theta +\frac{\mathrm{sin}⁡6\theta }{6}]}_{x/6}^{5\pi /6}=\frac{1}{2}[\frac{5\pi }{6}-\frac{\pi }{6}+\frac{1}{6}(0-0)]=\frac{1}{2}.\frac{4\pi }{6}=\frac{\pi }{3}$$=21​[θ+6sin⁡6θ​]x/65π/6​=21​[65π​−6π​+61​(0−0)]=21​.64π​=3π​