JEE Main 2022 · 24 June · Evening Shift · Mathematics · Trigonometric Equations
D. 777
Given,
cos(x+π3)cos(x−π3)=14cos2(2x)\cos \left(x + \dfrac{\pi}{3}\right)\cos \left(x - \dfrac{\pi}{3}\right) = \dfrac{1}{4}\cos^{2}\left(2x\right)cos(x+3π)cos(x−3π)=41cos2(2x)
⇒2cos(x+π3)cos(x−π3)=12cos2(2x)⇒ 2\cos \left(x + \dfrac{\pi}{3}\right)\cos \left(x - \dfrac{\pi}{3}\right) = \dfrac{1}{2}\cos^{2}\left(2x\right)⇒2cos(x+3π)cos(x−3π)=21cos2(2x)
⇒cos(2x)+cos(2π3)=12cos22x⇒ \cos \left(2x\right) + \cos \left(\dfrac{2\pi}{3}\right) = \dfrac{1}{2}\cos^{2}2x⇒cos(2x)+cos(32π)=21cos22x
⇒cos2x+(−12)=12cos22x⇒ \cos 2x + \left(- \dfrac{1}{2}\right) = \dfrac{1}{2}\cos^{2}2x⇒cos2x+(−21)=21cos22x
⇒cos22x−2cos2x+1=0⇒ \cos^{2}2x - 2\cos 2x + 1 = 0⇒cos22x−2cos2x+1=0
⇒(cos2x−1)2=0⇒ {\left(\cos 2x - 1\right)}^{2} = 0⇒(cos2x−1)2=0or cos2x=1\cos 2x = 1cos2x=1
⇒2x=−6π,−4π,−2π,0,2π,4π,6π⇒ 2x = - 6\pi, - 4\pi, - 2\pi, 0, 2\pi, 4\pi, 6\pi⇒2x=−6π,−4π,−2π,0,2π,4π,6π
So, x∈{−3π,−2π,−π,0,π,2π,3π}x \in \left\{- 3\pi, - 2\pi, - \pi, 0, \pi, 2\pi, 3\pi\right\}x∈{−3π,−2π,−π,0,π,2π,3π}
So total 777 solutions.