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Question Number 109493 by bemath last updated on 24/Aug/20

	Answered by Dwaipayan Shikari last updated on 24/Aug/20

	${c o s}^{2} 2 x + {c o s}^{2} 3 x + {c o s}^{2} 4 x = \frac{3}{2}$ $1 + c o s 4 x + 1 + c o s 6 x + 1 + c o s 8 x = 3$ $c o s 4 x + c o s 6 x + c o s 8 x = 0$ $2 c o s 6 x c o s 2 x + c o s 6 x = 0$ $2 c o s 6 x (c o s 2 x + \frac{1}{2}) = 0$ $2 c o s 6 x = 0 \Rightarrow 6 x = π k + \frac{π}{2} \Rightarrow x = \frac{π k}{6} + \frac{π}{12} (k \in Z)$ $o r c o s 2 x + \frac{1}{2} = 0$ $\Rightarrow c o s 2 x = - \frac{1}{2} = c o s \frac{2 π}{3} \Rightarrow 2 x = 2 π k \pm \frac{2 π}{3} \Rightarrow x = π k \pm \frac{π}{3}$
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