On-the-interval-of-0-2pi-solve-sin-6x-sin-2x-0-

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