JEE Main 2019 — Mathematics Trigonometry

Given ${\mathrm{sin}}^{2}2\theta +{\mathrm{cos}}^{4}2\theta =\frac{3}{4}$sin22θ+cos42θ=43​

Using ${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1,$sin2x+cos2x=1, we get

$$\Rightarrow 1-{\mathrm{cos}}^{2}2\theta +{\mathrm{cos}}^{4}2\theta =\frac{3}{4}$$⇒1−cos22θ+cos42θ=43​

Let ${\mathrm{cos}}^{2}2\theta =t$cos22θ=t

$$\Rightarrow {t}^{2}-t+\frac{1}{4}=0$$⇒t2−t+41​=0

$$\Rightarrow {(t-\frac{1}{2})}^{2}=0$$⇒(t−21​)2=0

$$\Rightarrow t=\frac{1}{2}$$⇒t=21​

$$\Rightarrow {\mathrm{cos}}^{2}2\theta =\frac{1}{2}$$⇒cos22θ=21​

$$\Rightarrow 2{\mathrm{cos}}^{2}2\theta -1=0$$⇒2cos22θ−1=0

Using $\mathrm{cos}2A=2{\mathrm{cos}}^{2}A-1,$cos2A=2cos2A−1, we get

$$\Rightarrow \mathrm{cos}4\theta =0$$⇒cos4θ=0

If $\mathrm{cos}x=0,$cosx=0, then $x=(2n+1)\frac{\pi }{2},n\in Z.$x=(2n+1)2π​,n∈Z.

$$\Rightarrow 4\theta =(2n+1)\frac{\pi }{2}$$⇒4θ=(2n+1)2π​

$$\Rightarrow \theta =(2n+1)\frac{\pi }{8}$$⇒θ=(2n+1)8π​

$$\Rightarrow \theta =\frac{\pi }{8},\frac{3\pi }{8}\in (0, \frac{\pi }{2})$$⇒θ=8π​,83π​∈(0,2π​)

Hence, the sum of the values of $\theta$θ is $\frac{\pi }{8}+\frac{3\pi }{8}=\frac{\pi }{2}.$8π​+83π​=2π​.