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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

The sum of all the solution of the equation θ ((π)/(3)+θ)...

The sum of all the solution of the equation cos⁡θcos⁡(π3+θ)cos⁡(π3−θ)=14,θ∈[0,6π]\cos \theta \cos \left(\frac{\pi}{3}+\theta\right) \cos \left(\frac{\pi}{3}-\theta\right)=\frac{1}{4}, \theta \in[0,6 \pi]

Options

  1. A15π15 \pi
  2. B30π30 \pi
  3. C100π3\frac{100 \pi}{3}
  4. DNone of these

Correct answer

B. 30π30 \pi

Step-by-step solution

<br/>2cos⁡θ[cos⁡120∘+cos⁡2θ]=1<br/>⇒2cos⁡θ(−12+2cos⁡2θ−1)=1<br/>⇒4cos⁡3θ−3cos⁡θ−1=0<br/>⇒cos⁡3θ=1=cos⁡0<br/>⇒3θ=2nπ or θ=2nπ3,n∈z<br/>\begin{aligned}<br />& 2 \cos \theta\left[\cos 120^{{}^{{}^{\circ}}}+\cos 2 \theta\right]=1 \\<br />\Rightarrow & 2 \cos \theta\left(-\frac{1}{2}+2 \cos ^2 \theta-1\right)=1 \\<br />\Rightarrow & 4 \cos ^3 \theta-3 \cos \theta-1=0 \\<br />\Rightarrow & \cos 3 \theta=1=\cos 0 \\<br />\Rightarrow & 3 \theta=2 n \pi \text { or } \theta=\frac{2 n \pi}{3}, n \in z<br />\end{aligned}
Given the values so that 2n2 n does not exceed 18 .
Hence, the sum =2π3∑19n=2π3×9(9+1)2=30π=\frac{2 \pi}{3} \sum_1^9 n=\frac{2 \pi}{3} \times \frac{9(9+1)}{2}=30 \pi
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