Question:
If \(6\sin^2x=3\cos^4x-\sin^2x\cos^2x\), then \(x=\)
If \(6\sin^2x=3\cos^4x-\sin^2x\cos^2x\), then \(x=\)
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Whenever higher powers of \(\sin x\) and \(\cos x\) appear, convert everything into one trigonometric ratio.
Updated On: Jun 17, 2026
- \(2n\pi\pm\dfrac{\pi}{3},\ \forall n\in\mathbb{Z}\)
- \(n\pi\pm\dfrac{\pi}{3},\ \forall n\in\mathbb{Z}\)
- \(n\pi\pm\dfrac{\pi}{6},\ \forall n\in\mathbb{Z}\)
- \(2n\pi\pm\dfrac{\pi}{4},\ \forall n\in\mathbb{Z}\)
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The Correct Option is C
Solution and Explanation
Concept:
Use trigonometric identities and convert the equation into a quadratic form.
Step 1: Write the given equation.
\[ 6\sin^2x=3\cos^4x-\sin^2x\cos^2x \] Using \[ \sin^2x=1-\cos^2x \] let \[ \cos^2x=t \] Then, \[ 6(1-t)=3t^2-(1-t)t \]
Step 2: Simplify the equation.
\[ 6-6t=3t^2-t+t^2 \] \[ 6-6t=4t^2-t \] \[ 4t^2+5t-6=0 \] \[ (4t-3)(t+2)=0 \] Thus, \[ t=\frac34 \] Since \(t=\cos^2x\), \[ \cos^2x=\frac34 \] \[ \cos x=\pm\frac{\sqrt3}{2} \]
Step 3: Find the general solution.
\[ x=n\pi\pm\frac{\pi}{6},\qquad n\in\mathbb{Z} \] Hence, \[ \boxed{x=n\pi\pm\frac{\pi}{6}} \]
Step 1: Write the given equation.
\[ 6\sin^2x=3\cos^4x-\sin^2x\cos^2x \] Using \[ \sin^2x=1-\cos^2x \] let \[ \cos^2x=t \] Then, \[ 6(1-t)=3t^2-(1-t)t \]
Step 2: Simplify the equation.
\[ 6-6t=3t^2-t+t^2 \] \[ 6-6t=4t^2-t \] \[ 4t^2+5t-6=0 \] \[ (4t-3)(t+2)=0 \] Thus, \[ t=\frac34 \] Since \(t=\cos^2x\), \[ \cos^2x=\frac34 \] \[ \cos x=\pm\frac{\sqrt3}{2} \]
Step 3: Find the general solution.
\[ x=n\pi\pm\frac{\pi}{6},\qquad n\in\mathbb{Z} \] Hence, \[ \boxed{x=n\pi\pm\frac{\pi}{6}} \]
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