JEE Main 2022 — Mathematics Trigonometry

Given,

$$2{\mathrm{sin}}^{2}\theta -\mathrm{cos}2\theta =0$$2sin2θ−cos2θ=0

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

$$\Rightarrow {\mathrm{sin}}^{2}\theta ={(\frac{1}{2})}^{2}$$⇒sin2θ=(21​)2

$$\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$$⇒θ=6π​,65π​,67π​,611π​

Also given, $2{\mathrm{cos}}^{2}\theta +3\mathrm{sin}\theta =0$2cos2θ+3sinθ=0

$$\Rightarrow 2{\mathrm{sin}}^{2}\theta -3\mathrm{sin}\theta -2=0$$⇒2sin2θ−3sinθ−2=0

$$\Rightarrow \mathrm{sin}\theta =-\frac{1}{2}$$⇒sinθ=−21​

$$\Rightarrow \theta =\frac{7\pi }{6},\frac{11\pi }{6}$$⇒θ=67π​,611π​

So, the common solution is

$$\theta =\frac{7\pi }{6},\frac{11\pi }{6}$$θ=67π​,611π​

Sum $=\frac{7\pi +11\pi }{6}=3\pi =k\pi$=67π+11π​=3π=kπ

$$\Rightarrow K=3$$⇒K=3