JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

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Analyzing +4 | -1

The solution of the equation

{\cos ^2}x - 2\cos x = 4\sin x - \sin \,\,2x\,\,\left( {0 \le x \le \pi } \right)

View answer and explanation

Correct option: C

\pi + {\tan ^{ - 1}}\left( { - \frac{1}{2}} \right)

Option D (

\pi + {\tan ^{ - 1}}\left( { - \frac{1}{2}} \right)

) is correct because by simplifying the given equation, we find that

\tan x = -1/2

and

x = \pi + {\tan ^{ - 1}}\left( { - \frac{1}{2}} \right)

in the given range.

More Questions from TRIGONOMETRIC EQUATIONS

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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