Question:
Statement I: If \( x\in(0,\frac{\pi}{2}) \) and \( cos~3x+cos~x=cos~2x \), then \( x=\frac{\pi}{4} \) or \( \frac{\pi}{3} \)
Statement - II: If \( sin~x~sin~2x=cos~x~cos~2x-1 \), then \( x=\frac{n\pi}{3},n\in Z \)
Which of the following options is correct?
Statement I: If \( x\in(0,\frac{\pi}{2}) \) and \( cos~3x+cos~x=cos~2x \), then \( x=\frac{\pi}{4} \) or \( \frac{\pi}{3} \)
Statement - II: If \( sin~x~sin~2x=cos~x~cos~2x-1 \), then \( x=\frac{n\pi}{3},n\in Z \)
Which of the following options is correct?
Statement - II: If \( sin~x~sin~2x=cos~x~cos~2x-1 \), then \( x=\frac{n\pi}{3},n\in Z \)
Which of the following options is correct?
Show Hint
Be cautious when choosing general solution forms. For example, \( cos~\theta = 1 \) uniquely maps to even multiples of \( \pi \) (\( 2n\pi \)), whereas \( cos^2\theta = 1 \) or general zero boundaries introduce steps of \( n\pi \).
Updated On: Jun 8, 2026
- I is true and II is true
- I is false and II is true
- I is true and II is false
- I is false and II is false
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
We solve both trigonometric equations independently using standard sum-to-product identities and basic trigonometric definitions.
Step 1: Evaluating Statement I.
Given equation: \( cos~3x + cos~x = cos~2x \). Apply the sum-to-product formula \( cos~A + cos~B = 2cos\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right) \): \[ 2cos~2x \cdot cos~x = cos~2x \implies cos~2x(2cos~x - 1) = 0 \] This yields two possible branches:
• \( cos~2x = 0 \implies 2x = \frac{\pi}{2} \implies x = \frac{\pi}{4} \) (lies inside the interval \( (0, \frac{\pi}{2}) \)).
• \( 2cos~x - 1 = 0 \implies cos~x = \frac{1}{2} \implies x = \frac{\pi}{3} \) (lies inside the interval \( (0, \frac{\pi}{2}) \)).
Hence, Statement I is completely true.
Step 2: Evaluating Statement II.
Given equation: \( sin~x~sin~2x = cos~x~cos~2x - 1 \). Rearranging the terms: \[ cos~x~cos~2x - sin~x~sin~2x = 1 \] Using the cosine addition formula \( cos(A+B) = cos~A~cos~B - sin~A~sin~B \): \[ cos(2x + x) = 1 \implies cos~3x = 1 \] The general solution for \( cos~\theta = 1 \) is \( \theta = 2n\pi \). Therefore: \[ 3x = 2n\pi \implies x = \frac{2n\pi}{3}, \quad n \in \mathbb{Z} \] The statement claims the solution is \( \frac{n\pi}{3} \), which includes extra invalid fractions. Hence, Statement II is false.
Step 1: Evaluating Statement I.
Given equation: \( cos~3x + cos~x = cos~2x \). Apply the sum-to-product formula \( cos~A + cos~B = 2cos\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right) \): \[ 2cos~2x \cdot cos~x = cos~2x \implies cos~2x(2cos~x - 1) = 0 \] This yields two possible branches:
• \( cos~2x = 0 \implies 2x = \frac{\pi}{2} \implies x = \frac{\pi}{4} \) (lies inside the interval \( (0, \frac{\pi}{2}) \)).
• \( 2cos~x - 1 = 0 \implies cos~x = \frac{1}{2} \implies x = \frac{\pi}{3} \) (lies inside the interval \( (0, \frac{\pi}{2}) \)).
Hence, Statement I is completely true.
Step 2: Evaluating Statement II.
Given equation: \( sin~x~sin~2x = cos~x~cos~2x - 1 \). Rearranging the terms: \[ cos~x~cos~2x - sin~x~sin~2x = 1 \] Using the cosine addition formula \( cos(A+B) = cos~A~cos~B - sin~A~sin~B \): \[ cos(2x + x) = 1 \implies cos~3x = 1 \] The general solution for \( cos~\theta = 1 \) is \( \theta = 2n\pi \). Therefore: \[ 3x = 2n\pi \implies x = \frac{2n\pi}{3}, \quad n \in \mathbb{Z} \] The statement claims the solution is \( \frac{n\pi}{3} \), which includes extra invalid fractions. Hence, Statement II is false.
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