The expression sin x + sqrt{3} cos x is maxim | vedclass

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(C) Let $f(x) = \sin x + \sqrt{3} \cos x$.To find the maximum value,we can rewrite the expression as $f(x) = 2 \left( \frac{1}{2} \sin x + \frac{\sqrt

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(C) Let $f(x) = \sin x + \sqrt{3} \cos x$.To find the maximum value,we can rewrite the expression as $f(x) = 2 \left( \frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x ight)$.$f(x) = 2 \left( \sin x \cos 60^\circ + \cos x \sin 60^\circ ight) = 2 \sin(x + 60^\circ)$.The function $\sin(x + 60^\circ)$ is maximum when the argument is $90^\circ$.$x + 60^\circ = 90^\circ \Rightarrow x = 30^\circ$.Alternatively,using derivatives:$f'(x) = \cos x - \sqrt{3} \sin x = 0 \Rightarrow 	an x = \frac{1}{\sqrt{3}} \Rightarrow x = 30^\circ$.At $x = 30^\circ$,$f''(x) = -\sin x - \sqrt{3} \cos x = -\sin 30^\circ - \sqrt{3} \cos 30^\circ = -\frac{1}{2} - \sqrt{3} \left( \frac{\sqrt{3}}{2} ight) = -\frac{1}{2} - \frac{3}{2} = -2 Since the second derivative is negative,the function is maximum at $x = 30^\circ$.

The expression $\sin x + \sqrt{3} \cos x$ is maximum when $x = \dots \, ^\circ$.

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