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?

• A. I is true and II is true
• B. I is false and II is true
• C. I is true and II is false
• D. I is false and II is false

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 \( n\pi \).

Answer 1

The Correct Option is C

Step 1: Tackle Statement I.
For $\cos 3x + \cos x = \cos 2x$, use the sum-to-product rule: $\cos 3x + \cos x = 2\cos 2x\cos x$. So $2\cos 2x\cos x = \cos 2x$.
Step 2: Factor it.
This gives $\cos 2x(2\cos x - 1) = 0$.
Step 3: Solve each branch in the interval.
$\cos 2x = 0$ gives $x = \tfrac{\pi}{4}$, and $\cos x = \tfrac{1}{2}$ gives $x = \tfrac{\pi}{3}$. Both lie in $(0,\tfrac{\pi}{2})$, so Statement I is true.
Step 4: Tackle Statement II.
For $\sin x\sin 2x = \cos x\cos 2x - 1$, move terms: $\cos x\cos 2x - \sin x\sin 2x = 1$.
Step 5: Use the cosine addition rule.
The left side is $\cos(2x+x) = \cos 3x$, so $\cos 3x = 1$. Then $3x = 2n\pi$, giving $x = \tfrac{2n\pi}{3}$.
Step 6: Compare with the claim.
Statement II claims $x = \tfrac{n\pi}{3}$, which adds extra wrong values. So II is false, while I is true. \[ \boxed{\text{I true, II false}} \]