The number of turning points of the curve f(x | vedclass

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(B) To find the turning points,we need to find the points where the derivative $f'(x) = 0$.Given $f(x) = 2 \cos x - \sin 2x$.Differentiating with resp

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$3$

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(B) To find the turning points,we need to find the points where the derivative $f'(x) = 0$.Given $f(x) = 2 \cos x - \sin 2x$.Differentiating with respect to $x$:$f'(x) = -2 \sin x - 2 \cos 2x$.Setting $f'(x) = 0$:$-2 \sin x - 2 \cos 2x = 0$$\sin x + \cos 2x = 0$.Using the identity $\cos 2x = 1 - 2 \sin^2 x$:$\sin x + 1 - 2 \sin^2 x = 0$$2 \sin^2 x - \sin x - 1 = 0$.Let $u = \sin x$,then $2u^2 - u - 1 = 0$.$(2u + 1)(u - 1) = 0$.So,$\sin x = 1$ or $\sin x = -1/2$.For $\sin x = 1$ in $[-\pi, \pi]$,$x = \pi/2$.For $\sin x = -1/2$ in $[-\pi, \pi]$,$x = -\pi/6$ and $x = -5\pi/6$.Thus,there are $3$ turning points in the given interval.

The number of turning points of the curve $f(x) = 2 \cos x - \sin 2x$ in the interval $[-\pi, \pi]$ is

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