The number of solutions of the equation sin x | vedclass

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(B) Given the equation $\sin x = \cos^{2} x$.Using the identity $\cos^{2} x = 1 - \sin^{2} x$,we get $\sin x = 1 - \sin^{2} x$.Rearranging the terms,w

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$4$

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(B) Given the equation $\sin x = \cos^{2} x$.Using the identity $\cos^{2} x = 1 - \sin^{2} x$,we get $\sin x = 1 - \sin^{2} x$.Rearranging the terms,we obtain the quadratic equation $\sin^{2} x + \sin x - 1 = 0$.Let $t = \sin x$. Then $t^{2} + t - 1 = 0$.Using the quadratic formula $t = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$,we find $t = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.Since $-1 \le \sin x \le 1$,we must have $\sin x = \frac{-1 + \sqrt{5}}{2} \approx 0.618$ (which is positive) or $\sin x = \frac{-1 - \sqrt{5}}{2} \approx -1.618$ (which is impossible).Thus,$\sin x = \frac{\sqrt{5} - 1}{2}$.In the interval $(0, 10)$,the value $10$ radians is approximately $10 / 3.14 \approx 3.18\pi$.Since $\sin x = k$ (where $0 In $(0, 2\pi)$,there are $2$ solutions.In $(2\pi, 3\pi)$,there are $0$ solutions (as $\sin x$ is negative).In $(3\pi, 3.18\pi)$,there are $2$ solutions.Total number of solutions is $2 + 2 = 4$.

The number of solutions of the equation $\sin x = \cos^{2} x$ in the interval $(0, 10)$ is

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