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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

The number of solutions of the equation 4 2 θ · 3 θ= θ , when 0 < θ...

The number of solutions of the equation 4cos⁡2θ⋅cos⁡3θ=sec⁡θ4 \cos 2 \theta \cdot \cos 3 \theta=\sec \theta, when 0<θ<π0 < \theta < \pi, is

Options

  1. A2
  2. B4
  3. C6
  4. D8

Correct answer

C. 6

Step-by-step solution

We have, 4cos⁡2θ⋅cos⁡3θ=sec⁡θ, (where 0<θ<π)⇒2(2cos⁡2θ⋅cos⁡3θ)=1cos⁡θ⇒2[cos⁡5θ+cos⁡(−θ)]=1cos⁡θ[∵2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)]⇒2[cos⁡5θ+cos⁡θ]=1cos⁡θ[∵cos⁡(−θ)=cos⁡θ]⇒2cos⁡5θ⋅cos⁡θ+2cos⁡2θ=1⇒(cos⁡6θ+cos⁡4θ)+(2cos⁡2θ−1)=0⇒cos⁡6θ+cos⁡4θ+cos⁡2θ=0⇒2cos⁡4θ⋅cos⁡2θ+cos⁡4θ=0⇒cos⁡4θ(2cos⁡2θ+1)=0⇒cos⁡4θ=0 and 2cos⁡2θ+1=0⇒4θ=(2n+1)π2,cos⁡2θ=−12⇒θ=(2n+1)π8,2θ=2π/3 or 4π3⇒θ=π8,3π8,5π8,7π8,π3,2π3\begin{aligned}& 4 \cos 2 \theta \cdot \cos3 \theta=\sec \theta \text {, (where }0 < \theta< \pi)\\ & \Rightarrow \quad 2(2 \cos 2 \theta \cdot \cos3 \theta)=\frac{1}{\cos \theta}\\ & \Rightarrow 2[\cos 5 \theta+\cos (-\theta)]=\frac{1}{\cos \theta}\\ & \quad \quad[\because 2 \cos A \cos B=\cos (A+B)+\cos (A-B)] \\ & \Rightarrow \quad 2[\cos 5 \theta+\cos \theta]=\frac{1}{\cos \theta} \quad[\because \cos (-\theta)=\cos \theta]\\ & \Rightarrow \quad 2 \cos 5 \theta \cdot \cos \theta+2 \cos ^2 \theta=1 \\ & \Rightarrow \quad(\cos 6 \theta+\cos4 \theta)+\left(2 \cos ^2 \theta-1\right)=0 \\ & \Rightarrow \quad \cos 6 \theta+\cos4 \theta+\cos2 \theta=0 \\ & \Rightarrow \quad 2 \cos 4 \theta \cdot \cos2 \theta+\cos4 \theta=0 \\ & \Rightarrow \quad \cos 4 \theta(2 \cos 2 \theta+1)=0 \\ & \Rightarrow \quad \cos 4 \theta=0 \text { and }2 \cos 2 \theta+1=0 \\ & \Rightarrow \quad 4 \theta=(2 n+1) \frac{\pi}{2}, \cos2 \theta=-\frac{1}{2}\\ & \Rightarrow \quad \theta=(2 n+1) \frac{\pi}{8},2 \theta=2 \pi/ 3 \text { or } \frac{4 \pi}{3}\\ & \Rightarrow \quad \theta=\frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}, \frac{\pi}{3}, \frac{2 \pi}{3} \end{aligned}So, number of solution of given equation is 6 .
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