Question:
If \( \theta \)1 and \( \theta \)2 be respectively the smallest and the largest values of \( \theta \) in (0, 2\( \pi \)) - {\( \pi \)} which satisfy the equation,
2cot2\( \theta \) - \( {5 \over {\sin \theta }} \) + 4 = 0, then
\( \int\limits_{{\theta _1}}^{{\theta _2}} {{{\cos }^2}3\theta d\theta } \) is equal to :
If \( \theta \)1 and \( \theta \)2 be respectively the smallest and the largest values of \( \theta \) in (0, 2\( \pi \)) - {\( \pi \)} which satisfy the equation,
2cot2\( \theta \) - \( {5 \over {\sin \theta }} \) + 4 = 0, then
\( \int\limits_{{\theta _1}}^{{\theta _2}} {{{\cos }^2}3\theta d\theta } \) is equal to :
2cot2\( \theta \) - \( {5 \over {\sin \theta }} \) + 4 = 0, then
\( \int\limits_{{\theta _1}}^{{\theta _2}} {{{\cos }^2}3\theta d\theta } \) is equal to :
Show Hint
Solve the trigonometric equation in \(\sin\theta\) to get \(\theta_1,\theta_2\), then integrate \(\cos^2 3\theta\) using the half-angle formula.
Updated On: Oct 31, 2026
- \( {\pi \over 9} \)
- \( {{2\pi } \over 3} \)
- \( {{\pi } \over 3} \)
- \( {\pi \over 3} + {1 \over 6} \)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: \(2\cot^2\theta-\dfrac5{\sin\theta}+4=0\). With \(s=\sin\theta\): \(\dfrac{2(1-s^2)}{s^2}-\dfrac5s+4=0\Rightarrow2s^2-5s+2=0\Rightarrow s=2\) or \(s=\tfrac12\).
Step 2: \(\sin\theta=\tfrac12\) in \((0,2\pi)\): \(\theta=\tfrac\pi6,\tfrac{5\pi}6\) (neither equals pi). So \(\theta_1=\tfrac\pi6,\;\theta_2=\tfrac{5\pi}6\).
Step 3: \(\int\cos^23\theta\,d\theta=\int\dfrac{1+\cos6\theta}2d\theta=\dfrac\theta2+\dfrac{\sin6\theta}{12}\).
Step 4: Between \(\tfrac\pi6\) and \(\tfrac{5\pi}6\): \(\dfrac12\cdot\dfrac{4\pi}6+\dfrac{\sin5\pi-\sin\pi}{12}=\dfrac\pi3+0\).
Answer: (C) \(\tfrac\pi3\)
Step 2: \(\sin\theta=\tfrac12\) in \((0,2\pi)\): \(\theta=\tfrac\pi6,\tfrac{5\pi}6\) (neither equals pi). So \(\theta_1=\tfrac\pi6,\;\theta_2=\tfrac{5\pi}6\).
Step 3: \(\int\cos^23\theta\,d\theta=\int\dfrac{1+\cos6\theta}2d\theta=\dfrac\theta2+\dfrac{\sin6\theta}{12}\).
Step 4: Between \(\tfrac\pi6\) and \(\tfrac{5\pi}6\): \(\dfrac12\cdot\dfrac{4\pi}6+\dfrac{\sin5\pi-\sin\pi}{12}=\dfrac\pi3+0\).
Answer: (C) \(\tfrac\pi3\)
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