Question:
The sum of all the solutions of the equation $\cos x · \cos(\frac\pi3+x) · \cos(\frac\pi3-x) = \frac14, x \in [0, 6π]$ is
The sum of all the solutions of the equation $\cos x · \cos(\frac\pi3+x) · \cos(\frac\pi3-x) = \frac14, x \in [0, 6π]$ is
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The sum of all the solutions of the equation $\cos x · \cos(π/3+x) · \cos(π/3-x) = \frac14, x \in [0, 6π]$ is
Updated On: Apr 15, 2026
- $15\pi$
- $30\pi$
- $\frac{110\pi}{3}$
- None of these
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Use the identity $\cos x \cos(60^\circ-x) \cos(60^\circ+x) = \frac{1}{4} \cos 3x$.
Step 2: Analysis
The equation becomes $\frac{1}{4} \cos 3x = \frac{1}{4}$, which means $\cos 3x = 1$.
Step 3: Evaluation
$\cos 3x = 1 \Rightarrow 3x = 2n\pi$, so $x = \frac{2n\pi}{3}$. For $x \in [0, 6\pi]$, $n$ can be $0, 1, 2, ..., 9$.
Step 4: Conclusion
Sum $= \frac{2\pi}{3} (0+1+2...+9) = \frac{2\pi}{3} \cdot 45 = 30\pi$.
Final Answer: (b)
Use the identity $\cos x \cos(60^\circ-x) \cos(60^\circ+x) = \frac{1}{4} \cos 3x$.
Step 2: Analysis
The equation becomes $\frac{1}{4} \cos 3x = \frac{1}{4}$, which means $\cos 3x = 1$.
Step 3: Evaluation
$\cos 3x = 1 \Rightarrow 3x = 2n\pi$, so $x = \frac{2n\pi}{3}$. For $x \in [0, 6\pi]$, $n$ can be $0, 1, 2, ..., 9$.
Step 4: Conclusion
Sum $= \frac{2\pi}{3} (0+1+2...+9) = \frac{2\pi}{3} \cdot 45 = 30\pi$.
Final Answer: (b)
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