JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Hard

1 of 1

Analyzing +4 | -1

If the equation

\cos \,\,\left( {\lambda \,\,\sin \,\,\theta } \right) = \sin \left( {\lambda \,\,\cos \,\,\theta } \right)

has a solution in

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

, then the smallest positive value of

\lambda

View answer and explanation

Correct option: D

\frac{\pi }{{2\,\sqrt 2 }}

Option D (

\frac{\pi }{{2\\,\sqrt 2 }}

) is correct because it represents the smallest positive value of

\lambda

that allows the given trigonometric equation to have a solution within the specified interval.

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