The sum of all the solution of the equation θ ((π)/(3)+θ) ((π)/(3)-θ)=(1)/(4), θ ∈[0,6 π]

Correct answer: B. 30 π aligned & 2 θ[ 120^ ^ + 2 θ]=1 & 2 θ(-(1)/(2)+2 ^2 θ-1)=1 & 4 ^3 θ-3 θ-1=0 & 3 θ=1= 0 & 3 θ=2 n π or θ=(2 n π)/(3),n ∈ z aligned Given the values so that 2 n does not exceed 18 . Hence, the sum = (2 π)/(3) ∑_1^9 n= (2 π)/(3) × (9(9+1))/(2)= 30 π

BITSAT 2022

Exam: BITSAT
Year: 2022
Subject: Mathematics
Chapter: Trigonometric Equations

BITSAT 2022 · Mathematics · Trigonometric Equations

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

The sum of all the solution of the equation

\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

A $15 \pi$
B $30 \pi$
C $\frac{100 \pi}{3}$
DNone of these

Correct answer

B. $30 \pi$

Step-by-step solution

\begin{aligned}& 2 \cos \theta\left[\cos120^{{}^{\circ}}+\cos 2 \theta\right]=1 \\ \Rightarrow & 2 \cos \theta\left(-\frac{1}{2}+2 \cos ^2 \theta-1\right)=1 \\ \Rightarrow & 4 \cos ^3 \theta-3 \cos \theta-1=0 \\ \Rightarrow & \cos 3 \theta=1=\cos 0 \\ \Rightarrow & 3 \theta=2 n \pi \text{\text{ or } } \theta=\frac{2 n \pi}{3},n \in z \end{aligned}

Given the values so that 2 n

does not exceed 18 . Hence, the sum =

\frac{2 \pi}{3}

\sum_1^9 n=

\frac{2 \pi}{3}

\times

\frac{9(9+1)}{2}=$30 \pi

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