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JEE Main 2024
Trigonometric Equations
Trigonometric Equations
Easy

Question

The number of solutions of sin2x+(2+2xx2)sinx3(x1)2=0\sin ^2 x+\left(2+2 x-x^2\right) \sin x-3(x-1)^2=0, where πxπ-\pi \leq x \leq \pi, is ________.

Answer: 2

Solution

sin2x+(3(x1)2)sinx3(x1)2=0sin2x+3sinx(x1)2sinx3(x1)2=0sinx(sinx+3)(x1)2)[sinx+3]=0\begin{aligned} & \sin ^2 x+\left(3-(x-1)^2\right) \sin x-3(x-1)^2=0 \\ & \sin ^2 x+3 \sin x-(x-1)^2 \sin x-3(x-1)^2=0 \\ & \left.\sin x(\sin x+3)-(x-1)^2\right)[\sin x+3]=0 \end{aligned} There are two intersections between this graph. So, Number of solution will be 2 .

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