The smallest positive $x$ satisfying the equation ${cos}^{3} 3 x + {cos}^{3} 5 x = 8 {cos}^{3} 4 x \cdot {cos}^{3} x$ is

15^{\circ}

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18^{\circ}

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{22.5}^{\circ}

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30^{\circ}

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Solution

The correct option is B $18^{\circ}$
${cos}^{3} 3 x + {cos}^{3} 5 x = 8 {cos}^{3} 4 x \cdot {cos}^{3} x$
$\Rightarrow {cos}^{3} 3 x + {cos}^{3} 5 x = (2 cos 4 x \cdot cos x)^{3}$
$\Rightarrow {cos}^{3} 3 x + {cos}^{3} 5 x = (cos 5 x + cos 3 x)^{3}$
$\Rightarrow {cos}^{3} 3 x + {cos}^{3} 5 x = {cos}^{3} 5 x + {cos}^{3} 3 x + 3 cos 5 x cos 3 x (cos 5 x + cos 3 x)$
$\Rightarrow 3 cos 5 x cos 3 x (cos 5 x + cos 3 x) = 0$
Using $cos C + cos D = 2 cos (\frac{C + D}{2}) cos (\frac{C - D}{2}),$ we get
$(3 cos 5 x cos 3 x) (2 cos 4 x \cdot cos x) = 0$
$\Rightarrow cos x \cdot cos 3 x \cdot cos 4 x \cdot cos 5 x = 0$
$\Rightarrow x = (2 n + 1) \frac{π}{2}, (2 n + 1) \frac{π}{6}, (2 n + 1) \frac{π}{8}, (2 n + 1) \frac{π}{10}$ where $n \in Z$
Put $n = 0$
$x = \frac{π}{2}, \frac{π}{6}, \frac{π}{8}, \frac{π}{10}$
$\Rightarrow$ Smallest +ve values of $x$ is $\frac{π}{10}$ i.e., $18^{\circ}$

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