The largest positive solution of $1 + {\sin ^4}x = {\cos ^2}3x$ in $\left[ { - \,5\pi /2,\;\,5\pi /2} \right]$ is

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Analyzing +4 | -1

The largest positive solution of

1 + {\sin ^4}x = {\cos ^2}3x

\left[ { - \,5\pi /2,\;\,5\pi /2} \right]

View answer and explanation

Correct option: B

2\;\pi

Option B (

2\;\pi

) is correct because it is the largest positive solution within the given interval that satisfies both sin(3x) = 0 and sin(x) = 0.

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\sin \,\,x\,\,\;cos\,\,3x = \sin \,\,3x\,\,\cos \,\,5x

\left[ {0,\;\,\frac{\pi }{2}} \right]

\left( {x,\,y} \right)

x,\,\,y \in \left[ {0,10} \right]

\left( {\sqrt {{{\sin }^2} - \sin \,\,x + \frac{1}{2}} } \right){2^{{{\sec }^2}y}}\,\, \cdot \,\, \le 1

\mathrm{X}

\theta \rightarrow \frac{\pi}{2}

\sin x+\cos x=\frac{\sqrt{7}}{2}

x \in\left[0, \frac{\pi}{4}\right]

\tan \frac{x}{2}

\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}

0 < x < \pi

\cos x+\sin x=\frac{1}{2}

\tan x