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Question Number 149031 by puissant last updated on 02/Aug/21

	Answered by EDWIN88 last updated on 02/Aug/21
	$⇒((1+cos 2x+1+cos 4x+1+cos 6x)/2)=1 ⇒cos 6x+cos 4x+cos 2x=−1 ⇒2cos 4x cos 2x+cos 4x=−1 ⇒cos 4x(2cos 2x+1)=−1 ⇒(2cos^2 2x−1)(2cos 2x+1)=−1 let cos 2x=u ⇒(2u^2 −1)(2u+1)=−1 ⇒4u^3 +2u^2 −2u=0 ⇒2u(2u^2 +u−1)=0 ⇒2u(2u−1)(u+1)=0 { ((u=0⇒cos 2x=0; 2x=±(π/2)+2kπ ; x=±(π/4)+kπ)),((u=(1/2)⇒cos 2x=(1/2); 2x=± (π/3)+2kπ ; x=±(π/6)+kπ )),((u=−1⇒cos 2x=−1; 2x=±π+2kπ ; x=±(π/2)+kπ)) :}$
	$\Rightarrow \frac{1 + \cos 2 x + 1 + \cos 4 x + 1 + \cos 6 x}{2} = 1$ $\Rightarrow \cos 6 x + \cos 4 x + \cos 2 x = - 1$ $\Rightarrow 2 \cos 4 x \cos 2 x + \cos 4 x = - 1$ $\Rightarrow \cos 4 x (2 \cos 2 x + 1) = - 1$ $\Rightarrow (2 \cos^{2} 2 x - 1) (2 \cos 2 x + 1) = - 1$ $l e t \cos 2 x = u$ $\Rightarrow (2 u^{2} - 1) (2 u + 1) = - 1$ $\Rightarrow 4 u^{3} + 2 u^{2} - 2 u = 0$ $\Rightarrow 2 u (2 u^{2} + u - 1) = 0$ $\Rightarrow 2 u (2 u - 1) (u + 1) = 0$ ${\begin{cases} u = 0 \Rightarrow \cos 2 x = 0; 2 x = \pm \frac{π}{2} + 2 k π; x = \pm \frac{π}{4} + k π \\ u = \frac{1}{2} \Rightarrow \cos 2 x = \frac{1}{2}; 2 x = \pm \frac{π}{3} + 2 k π; x = \pm \frac{π}{6} + k π \\ u = - 1 \Rightarrow \cos 2 x = - 1; 2 x = \pm π + 2 k π; x = \pm \frac{π}{2} + k π \end{cases}$
	Commented by puissant last updated on 02/Aug/21

	$thanks . . .$
	Answered by MJS_new last updated on 02/Aug/21

	$use x = \arccos c$ $\Rightarrow$ $c^{2} + {(2 c^{2} - 1)}^{2} + c^{2} {(4 c^{2} - 3)}^{2} = 1$ $\Rightarrow$ $c^{6} - \frac{5}{4} c^{4} + \frac{3}{8} c^{2} = 0$ $c^{2} (c^{2} - \frac{1}{2}) (c^{2} - \frac{3}{4}) = 0$ ${it}^{'} s easy now$
	Commented by puissant last updated on 02/Aug/21

	$thanks . .$
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