The number of solutions x of the equation sin | vedclass

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(C) Given equation: $\sin(x+x^2) - \sin(x^2) = \sin x$Using the identity $\sin A - \sin B = 2 \sin(\frac{A-B}{2}) \cos(\frac{A+B}{2})$,we get:$2 \sin(

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$2$

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(C) Given equation: $\sin(x+x^2) - \sin(x^2) = \sin x$Using the identity $\sin A - \sin B = 2 \sin(\frac{A-B}{2}) \cos(\frac{A+B}{2})$,we get:$2 \sin(\frac{x}{2}) \cos(\frac{2x^2+x}{2}) = \sin x$Since $\sin x = 2 \sin(\frac{x}{2}) \cos(\frac{x}{2})$,the equation becomes:$2 \sin(\frac{x}{2}) \cos(\frac{2x^2+x}{2}) = 2 \sin(\frac{x}{2}) \cos(\frac{x}{2})$This implies $2 \sin(\frac{x}{2}) [\cos(\frac{2x^2+x}{2}) - \cos(\frac{x}{2})] = 0$Case $1$: $\sin(\frac{x}{2}) = 0$ $\Rightarrow \frac{x}{2} = n\pi$ $\Rightarrow x = 2n\pi$. For $n=1$,$x=2\pi \approx 6.28$,which is outside $[2, 3]$.Case $2$: $\cos(\frac{2x^2+x}{2}) = \cos(\frac{x}{2}) \Rightarrow \frac{2x^2+x}{2} = 2k\pi \pm \frac{x}{2}$Subcase $2a$: $x^2 + \frac{x}{2} = 2k\pi + \frac{x}{2} \Rightarrow x^2 = 2k\pi$. For $k=1$,$x = \sqrt{2\pi} \approx \sqrt{6.28} \approx 2.506$ (in $[2, 3]$).Subcase $2b$: $x^2 + \frac{x}{2} = 2k\pi - \frac{x}{2} \Rightarrow x^2 + x - 2k\pi = 0$. For $k=1$,$x^2 + x - 2\pi = 0$. Roots are $x = \frac{-1 \pm \sqrt{1 + 8\pi}}{2}$. Taking positive root,$x = \frac{-1 + \sqrt{1 + 25.13}}{2} \approx \frac{-1 + 5.11}{2} \approx 2.055$ (in $[2, 3]$).Thus,there are $2$ solutions.

The number of solutions $x$ of the equation $\sin(x+x^2) - \sin(x^2) = \sin x$ in the interval $[2, 3]$ is

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