JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Medium

1 of 1

Application +4 | -1

The solutions of the equation

4\,\,{\cos ^2}\,\,x + 6\,\,{\sin ^2}\,\,x = 5

are

View answer and explanation

Correct option: A

x = n\pi \,\, \pm \,\,\frac{\pi }{4}

Option A (

x = n\pi \, \pm \, \frac{\pi }{4}

) is correct because solving the trigonometric equation

4{\cos ^2}x + 6{\sin ^2}x = 5

leads to

\sin x = \pm \frac{1}{\sqrt{2}}

, which gives the general solution

x = n\pi \pm \frac{\pi}{4}

More Questions from TRIGONOMETRIC EQUATIONS

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