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The number of solutions of the equation \sin x+2 \sin 2 x=3+\sin 3 x in the interval [0, \pi] is

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

best_answer

\begin{aligned} & \sin 3 x-\sin x-2 \sin 2 x+3=0 \\ \\& \Rightarrow \quad 2 \cos 2 x \sin x-4 \sin x \cos x+3=0 \\ \\& \Rightarrow \quad \sin x(2 \cos 2 x-4 \cos x)+3=0 \\ \\& \Rightarrow \sin x\left\{2\left(2 \cos ^2 x-1\right)-4 \cos x\right\}+3=0 \\ \\& \Rightarrow \quad \sin x\left\{4 \cos ^2 x-4 \cos x-2\right\}+3=0 \\ & \end{aligned}

\begin{array}{lc} \Rightarrow & \sin x\left\{(2 \cos x-1)^2-3\right\}+3=0 \\ \\\Rightarrow & \sin x(2 \cos x-1)^2+3(1-\sin x)=0 \\ \\\because & 0 \leq x \leq \pi \\ \\\Rightarrow & 0 \leq \sin x \leq 1 \end{array}

Then ,               1-\sin x \geq 0

Also \sin x(2 \cos x-1)^2 \geq 0

Which is possible only if \sin x=1

Then                                \cos x=0

\Rightarrow \quad \sin x(2 \cos x-1)^2=1 \neq 0

ie, no solution 

\therefore No. of solutions = 0 

Posted by

Gaurav

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