On the interval of [0,2π] solve sin 6x +sin 2x=0


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Question Number 107977 by anonymous last updated on 13/Aug/20

$O n t h e i n t e r v a l o f [0, 2 π] s o l v e$ $\sin 6 x + \sin 2 x = 0$
	Answered by bemath last updated on 13/Aug/20

	$\frac{B e \frac{M a t h}{♡}}{j o s s}$ $\Rightarrow 2 \sin 4 x \cos 2 x = 0$ ${\begin{cases} \sin 4 x = 0 \Rightarrow x = \frac{π}{2} k \to x = 0, \frac{π}{2}, \frac{3 π}{2}, 2 π \\ \cos 2 x = 0 \Rightarrow x = \pm \frac{π}{4} + m . π \end{cases}$
	Commented by anonymous last updated on 13/Aug/20

	$GODBLESS YOU$
	Answered by mr W last updated on 13/Aug/20

	$\sin 6 x = - \sin 2 x$ $\Rightarrow 6 x = (2 k + 1) π + 2 x \Rightarrow x = \frac{(2 k + 1) π}{4}$ $\Rightarrow 6 x = 2 k π - 2 x \Rightarrow x = \frac{k π}{4}$ $w i t h i n [0, 2 π] :$ $x = 0, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, π, \frac{5 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}, 2 π$
	Commented by anonymous last updated on 13/Aug/20

	$GOD BLESS YOU$
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