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Question

(A) Given equation: $\cos x + \cos 4x + \cos 2x + \cos 3x = 0$Using the sum-to-product formula $\cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \c

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Given equation: $\cos x + \cos 4x + \cos 2x + \cos 3x = 0$Using the sum-to-product formula $\cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$:$2 \cos \left(\frac{5x}{2}\right) \cos \left(\frac{3x}{2}\right) + 2 \cos \left(\frac{5x}{2}\right) \cos \left(\frac{x}{2}\right) = 0$$2 \cos \left(\frac{5x}{2}\right) [\cos \left(\frac{3x}{2}\right) + \cos \left(\frac{x}{2}\right)] = 0$Using $\cos C + \cos D = 2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$:$2 \cos \left(\frac{5x}{2}\right) [2 \cos x \cos \left(\frac{x}{2}\right)] = 0$$4 \cos \left(\frac{5x}{2}\right) \cos x \cos \left(\frac{x}{2}\right) = 0$Case $1$: $\cos \left(\frac{5x}{2}\right) = 0$ $\Rightarrow \frac{5x}{2} = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}$ $\Rightarrow x = \frac{\pi}{5}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{9\pi}{5}$Case $2$: $\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}$Case $3$: $\cos \left(\frac{x}{2}\right) = 0$ $\Rightarrow \frac{x}{2} = \frac{\pi}{2}$ $\Rightarrow x = \pi$ (already included)The distinct values are $\frac{\pi}{5}, \frac{\pi}{2}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{3\pi}{2}, \frac{9\pi}{5}$.Total number of values is $7$.

If $0 \le x < 2\pi$,then the number of real values of $x$ which satisfy the equation $\cos x + \cos 2x + \cos 3x + \cos 4x = 0$ is . . .

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