The number of solutions of sin ^2 x + (2 + 2x | vedclass

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(C) Given equation: $\sin ^2 x + (2 + 2x - x^2) \sin x - 3(x - 1)^2 = 0$Rewrite the middle term: $2 + 2x - x^2 = 3 - (x^2 - 2x + 1) = 3 - (x - 1)^2$Le

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$2$

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(C) Given equation: $\sin ^2 x + (2 + 2x - x^2) \sin x - 3(x - 1)^2 = 0$Rewrite the middle term: $2 + 2x - x^2 = 3 - (x^2 - 2x + 1) = 3 - (x - 1)^2$Let $u = \sin x$ and $v = (x - 1)^2$. The equation becomes $u^2 + (3 - v)u - 3v = 0$.Factoring the quadratic: $u^2 + 3u - vu - 3v = 0 \implies u(u + 3) - v(u + 3) = 0 \implies (u - v)(u + 3) = 0$.This gives two cases: $\sin x = -3$ (which is impossible as $-1 \leq \sin x \leq 1$) or $\sin x = (x - 1)^2$.We need to find the number of solutions for $\sin x = (x - 1)^2$ in the interval $[-\pi, \pi]$.Let $f(x) = \sin x$ and $g(x) = (x - 1)^2$. $g(x) = 0$ at $x = 1$. Since $1 At $x = 1$,$f(1) = \sin(1) \approx 0.84$ and $g(1) = 0$. Since $f(1) > g(1)$,and $g(x)$ is a parabola opening upwards with vertex at $(1, 0)$,we check intersection points.For $x > 1$,$g(x)$ increases rapidly. At $x = 1 + \sqrt{1} = 2$,$g(2) = 1$ and $\sin(2) Graphically,the curve $y = \sin x$ and $y = (x - 1)^2$ intersect at $2$ points in the given interval $[-\pi, \pi]$.

The number of solutions of $\sin ^2 x + (2 + 2x - x^2) \sin x - 3(x - 1)^2 = 0$,where $-\pi \leq x \leq \pi$,is....................

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