The solution set of equation cos^2 x+cos^2 2x+cos^2 3x = 1 on 0≤x≤2π


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Question Number 137382 by bramlexs22 last updated on 02/Apr/21

$T h e s o l u t i o n s e t o f e q u a t i o n$ $\cos^{2} x + \cos^{2} 2 x + \cos^{2} 3 x = 1$ $o n 0 ⩽ x ⩽ 2 π$
Commented by MJS_new last updated on 03/Apr/21

$I get for 0 ⩽ x < 2 π$ $x \in {\frac{π}{6}, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{5 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}, \frac{11 π}{6}}$
	Answered by EDWIN88 last updated on 02/Apr/21
	$cos^2 x+cos^2 2x+cos^2 3x = 1 cos^2 2x + (cos 3x+cos x)^2 −2cos 3x cos x = 1 cos^2 2x+4cos^2 2x cos^2 x−(cos 4x+cos 2x)=1 cos^2 2x+4cos^2 2x cos^2 x−(2cos^2 2x−1+cos 2x)=1 cos^2 2x+4cos^2 2x cos^2 x−2cos^2 2x−cos 2x = 0 4cos^2 2x cos^2 x−cos^2 2x−cos 2x = 0 cos 2x(4cos 2x cos^2 x−cos 2x−1) = 0 (∗) cos 2x = 0 x = ± (π/4) + nπ (∗∗) 2cos 2x(1+cos 2x)−cos 2x−1=0 2cos^2 2x+cos 2x−1 = 0 (2cos 2x−1)(cos 2x+1)=0 { ((cos 2x=(1/2) ; x = ± (π/6) + nπ)),((cos 2x =− 1 ; x=±(π/2)+nπ )) :}$
	$\cos^{2} x + \cos^{2} 2 x + \cos^{2} 3 x = 1$ $\cos^{2} 2 x + {(\cos 3 x + \cos x)}^{2} - 2 \cos 3 x \cos x = 1$ $\cos^{2} 2 x + 4 \cos^{2} 2 x \cos^{2} x - (\cos 4 x + \cos 2 x) = 1$ $\cos^{2} 2 x + 4 \cos^{2} 2 x \cos^{2} x - (2 \cos^{2} 2 x - 1 + \cos 2 x) = 1$ $\cos^{2} 2 x + 4 \cos^{2} 2 x \cos^{2} x - 2 \cos^{2} 2 x - \cos 2 x = 0$ $4 \cos^{2} 2 x \cos^{2} x - \cos^{2} 2 x - \cos 2 x = 0$ $\cos 2 x (4 \cos 2 x \cos^{2} x - \cos 2 x - 1) = 0$ $() \cos 2 x = 0 x = \pm \frac{π}{4} + n π$ $( *) 2 \cos 2 x (1 + \cos 2 x) - \cos 2 x - 1 = 0$ $2 \cos^{2} 2 x + \cos 2 x - 1 = 0$ $(2 \cos 2 x - 1) (\cos 2 x + 1) = 0$ ${\begin{cases} \cos 2 x = \frac{1}{2}; x = \pm \frac{π}{6} + n π \\ \cos 2 x = - 1; x = \pm \frac{π}{2} + n π \end{cases}$
	Answered by liberty last updated on 02/Apr/21
	$⇔ ((1+cos 2x)/2) + ((1+cos 4x)/2) = 1−cos^2 3x ((2+cos 2x+cos 4x)/2) = ((1−cos 6x)/2) cos 6x+cos 4x+cos 2x +1 = 0 cos 4x+2cos 4x cos 2x +1 = 0 2cos^2 2x−1+2cos 4x cos 2x+1 = 0 2cos^2 2x+2cos 4x cos 2x = 0 2cos 2x(cos 2x+cos 4x)=0 2cos 2x(2cos 3x cos x)=0 { ((cos 2x = 0⇒x=± (π/4)+nπ)),((cos 3x=0⇒x=±(π/6)+((2nπ)/3))),((cos x=0⇒x=± (π/2)+2nπ)) :}$
	$\Leftrightarrow \frac{1 + \cos 2 x}{2} + \frac{1 + \cos 4 x}{2} = 1 - \cos^{2} 3 x$ $\frac{2 + \cos 2 x + \cos 4 x}{2} = \frac{1 - \cos 6 x}{2}$ $\cos 6 x + \cos 4 x + \cos 2 x + 1 = 0$ $\cos 4 x + 2 \cos 4 x \cos 2 x + 1 = 0$ $2 \cos^{2} 2 x - 1 + 2 \cos 4 x \cos 2 x + 1 = 0$ $2 \cos^{2} 2 x + 2 \cos 4 x \cos 2 x = 0$ $2 \cos 2 x (\cos 2 x + \cos 4 x) = 0$ $2 \cos 2 x (2 \cos 3 x \cos x) = 0$ ${\begin{cases} \cos 2 x = 0 \Rightarrow x = \pm \frac{π}{4} + n π \\ \cos 3 x = 0 \Rightarrow x = \pm \frac{π}{6} + \frac{2 n π}{3} \\ \cos x = 0 \Rightarrow x = \pm \frac{π}{2} + 2 n π \end{cases}$
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