If $\sec \,\,x\,\,\cos \,\,5x + 1 = 0,{\rm{\;}}\,\,{\rm{where\;}}\,\,0 < x < 2\pi ,\;\,{\rm{then}}\,\,\;x$ is equal to

$\frac{\pi }{5},\,\,\frac{\pi }{4}$

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

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Application +4 | -1

\sec \,\,x\,\,\cos \,\,5x + 1 = 0,\mathrm{\;}\,\,\mathrm{where\;}\,\,0 < x < 2\pi ,\;\,\mathrm{then}\,\,\;x

is equal to

View answer and explanation

Correct option: C

\frac{\pi }{4}

Option C,

\frac{\pi }{4}

, is correct because it is the only solution that satisfies the given equation

\sec \,x\,\cos \,5x + 1 = 0

within the interval

0 < x < 2\pi

Practice All Chapter MCQs →

\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