The total number of solutions of the equation cosx.cos2x.cos3x= $\frac{1}{4}$ in [0, $π$ ] is

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Solution

The correct option is B
6

$c o s x . c o s 2 x . c o s 3 x = \frac{1}{4} \Rightarrow 4 (c o s 3 x . c o s x) c o s 2 x = 1 \Rightarrow 2 (c o s 4 x + c o s 2 x) c o s 2 x = 1 \Rightarrow 2 (2 c o s^{2} 2 x - 1 + c o s 2 x) c o s 2 x = 1 \Rightarrow 4 c o s^{3} 2 x + 2 c o s^{2} 2 x - 2 c o s x - 1 = 0 \Rightarrow 2 c o s^{2} 2 x (2 c o s 2 x + 1) - (2 c o s 2 x + 1) = 0 \Rightarrow (2 c o s 2 x + 1) (2 c o s^{2} 2 x + 1) = 0 \Rightarrow c o s 4 x . (2 c o s 2 x + 1) = 0 c o s 4 x = 0, 2 c o s 2 x + 1 = 0 4 x (2 n_{1} + 1) \frac{π}{2}, n_{1} \in Z c o s 2 x = \frac{- 1}{2} x = (2 n_{1} + 1) \frac{π}{8}, n_{1} \in Z 2 x = 2 n_{2} π \pm \frac{2 π}{3}, n_{2} \in Z ∴ x = \frac{π}{8}, \frac{π}{8}, \frac{5 π}{7}, \frac{7 π}{8} ∴ x = \frac{π}{3}, \frac{2 π}{3}$

$∴$ The equation has 6 solutions.

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