Solve the equation: cos(6x)−cos(4x)=4y^2 +4y+3


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Question Number 145109 by mathdanisur last updated on 02/Jul/21

$S o l v e t h e e q u a t i o n :$ $c o s (6 x) - c o s (4 x) = 4 y^{2} + 4 y + 3$
	Answered by mitica last updated on 02/Jul/21

	$c o s 6 x - c o s 4 x ⩽ 1 - (- 1) = 2 \Rightarrow$ $4 y^{2} + 4 y + 3 ⩽ 2 \Rightarrow {(2 y + 1)}^{2} ⩽ 0$ $\Rightarrow y = \frac{- 1}{2}$ $c o s 4 x = - 1 \Rightarrow c o s 2 x = \frac{1 + c o s 4 x}{2} = 0$ $c o s 6 x = 1 \Rightarrow 4 {c o s}^{3} 2 x - 3 c o s 2 x = 1 \Rightarrow 0 = 1 \Rightarrow x \in ϕ$
	Commented by mathdanisur last updated on 02/Jul/21

	$a l o t c o o l S e r, t h a n k s$
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