JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

1 of 1

Analyzing +4 | -1

The number of solutions of the pair of equations

2{\sin ^2}\theta - \cos \,\,2\theta = 0

and

2{\cos ^2}\theta - 3\sin \,\,\theta = 0

in the interval

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

View answer and explanation

Correct option: C

Two

Option C (Two) is correct because solving the given equations yields

\sin \theta = \frac{1}{2}

, which has two solutions in the interval

[0, 2\pi]

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

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

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\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]

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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