If $\sin ^6 x=1+\cos ^4 3 x$, then $x$ is equal to

$(2 n-1) \frac{\pi}{4}, n \in I$

$(2 n+1) \frac{3 \pi}{4}, n \in I$

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Analyzing +4 | -1

\sin ^6 x=1+\cos ^4 3 x

, then

x

is equal to

View answer and explanation

Correct option: C

(2 n+1) \frac{\pi}{2}, n \in I

Option C (

(2 n+1) \frac{\pi}{2}, n \in I

) is the correct solution because it satisfies both

\sin^6 x = 1

and

\cos^4 3x = 0

Practice All Chapter MCQs →

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\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