JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

1 + \sin \,\,\theta + {\sin ^2}\,\,\theta + ...\infty = 4 + 2\,\sqrt 3 ,\,\;0 < \theta < \pi ,\;\,\theta \ne \frac{\pi }{2}

, then

View answer and explanation

Correct option: D

\theta = \frac{\pi }{3}

\theta = \frac{{2\pi }}{3}

The correct answer is D (

\theta = \frac{\pi }{3}

\theta = \frac{{2\pi }}{3}

) because these are the solutions to

\sin \theta = \frac{\sqrt{3}}{2}

in the given interval.

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

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\mathrm{X}

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

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\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