The set of all $x$ in $(-\pi, \pi)$ satisfying $|4 \sin x-1| < \sqrt{5}$ is given by

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

The set of all

x

(-\pi, \pi)

satisfying

|4 \sin x-1| < \sqrt{5}

is given by

View answer and explanation

Correct option: B

\left(\frac{\pi}{10}, \frac{3 \pi}{10}\right)

Option B

\left(\frac{\pi}{10}, \frac{3 \pi}{10}\right)

is correct because solving the inequality

|4 \sin x - 1| < \sqrt{5}

for

x

in the interval

(-\pi, \pi)

yields this solution set.

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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