Discussion
Let \(f(x) = (cos^{2}x)(a + \cos x)\).If \(f'\left(\frac{\pi}{3}\right)\)=0 then the value of \(a\) is equal to
Answer: C
\(a = \) \(\frac{-3}{4}\).
By the product rule:
\[ \begin{aligned} f'(x) &= -2\cos x\sin x\,(a + \cos x) - \cos^{2}x\sin x \\ &= -\sin x\cos x\,(2a + 3\cos x) \end{aligned} \]
At \(\frac{\pi}{3}\), \(\sin x\cos x \neq 0\), so \(2a + 3\cos\frac{\pi}{3} = 2a + \frac{3}{2} = 0\), giving \(a = -\frac{3}{4}\). \(\frac{-3}{2}\) forgets to divide by 2, and \(\frac{3}{4}\) has the sign wrong.
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