solve for x cos^2 x − cos^2 2x = cos^2 4x − cos^2 3x


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Question Number 153803 by weltr last updated on 10/Sep/21

$s o l v e f o r x$ ${c o s}^{2} x - {c o s}^{2} 2 x = {c o s}^{2} 4 x - {c o s}^{2} 3 x$
Commented by Ar Brandon last updated on 10/Sep/21

$x = 0 [π] is a solution$
	Answered by puissant last updated on 10/Sep/21

	$\Rightarrow 1 + c o s 2 x - 1 - c o s 4 x = 1 + c o s 8 x - 1 - c o s 6 x$ $\Rightarrow c o s 2 x - c o s 4 x + c o s 6 x - c o s 8 x = 0$ $\Rightarrow (c o s 2 x + c o s 6 x) - (c o s 4 x + c o s 8 x) = 0$ $\Rightarrow 2 c o s 2 x c o s 4 x - 2 c o s 2 x c o s 6 x = 0$ $\Rightarrow c o s 2 x (c o s 4 x - c o s 6 x) = 0$ $c o s 2 x = 0 o r c o s 4 x - c o s 6 x = 0$ $\to c o s 2 x = 0 \Rightarrow 2 x = \frac{π}{2} + k π, k \in Z$ $\Rightarrow x = \frac{π}{4} + \frac{k π}{2}, k \in Z . .$ $\to c o s 4 x - c o s 6 x = 0$ $\Rightarrow - 2 s i n (\frac{4 x - 6 x}{2}) s i n (\frac{4 x + 6 x}{2}) = 0$ $\Rightarrow 2 s i n x s i n 5 x = 0$ $\Rightarrow s i n x = 0 o r s i n 5 x = 0$ $\Rightarrow x = k π o r x = \frac{k π}{5}, k \in Z . .$ $∴∵ {x = \frac{π}{4} + \frac{k π}{2} o r x = k π o r x = \frac{k π}{5}; k \in Z} . .$
	Commented by weltr last updated on 10/Sep/21

	$t h a n k y o u$
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