Question Number 107977 by anonymous last updated on 13/Aug/20
![On the interval of [0,2π] solve sin 6x +sin 2x=0](Q107977.png)
$${On}\:{the}\:{interval}\:{of}\:\left[\mathrm{0},\mathrm{2}\pi\right]\:{solve} \\ $$$$\mathrm{sin}\:\mathrm{6}{x}\:+\mathrm{sin}\:\mathrm{2}{x}=\mathrm{0} \\ $$$$ \\ $$
Answered by bemath last updated on 13/Aug/20
$$\:\:\frac{\mathcal{B}{e}\frac{\mathcal{M}{ath}}{\heartsuit}}{{joss}} \\ $$$$\Rightarrow\:\mathrm{2sin}\:\mathrm{4}{x}\:\mathrm{cos}\:\mathrm{2}{x}\:=\:\mathrm{0}\: \\ $$$$\begin{cases}{\mathrm{sin}\:\mathrm{4}{x}\:=\:\mathrm{0}\:\Rightarrow{x}\:=\:\frac{\pi}{\mathrm{2}}{k}\:\rightarrow{x}=\mathrm{0},\frac{\pi}{\mathrm{2}},\frac{\mathrm{3}\pi}{\mathrm{2}},\mathrm{2}\pi}\\{\mathrm{cos}\:\mathrm{2}{x}\:=\:\mathrm{0}\Rightarrow{x}=\pm\frac{\pi}{\mathrm{4}}+{m}.\pi\:}\end{cases} \\ $$
Commented by anonymous last updated on 13/Aug/20
$$\mathbb{GODBLESS}\:\mathbb{YOU}\: \\ $$
Answered by mr W last updated on 13/Aug/20
![sin 6x=−sin 2x ⇒6x=(2k+1)π+2x ⇒x=(((2k+1)π)/4) ⇒6x=2kπ−2x ⇒x=((kπ)/4) within [0,2π]: x=0,(π/4),(π/2),((3π)/4),π,((5π)/4),((3π)/2),((7π)/4),2π](Q107980.png)
$$\mathrm{sin}\:\mathrm{6}{x}=−\mathrm{sin}\:\mathrm{2}{x} \\ $$$$\Rightarrow\mathrm{6}{x}=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi+\mathrm{2}{x}\:\Rightarrow{x}=\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{6}{x}=\mathrm{2}{k}\pi−\mathrm{2}{x}\:\Rightarrow{x}=\frac{{k}\pi}{\mathrm{4}} \\ $$$${within}\:\left[\mathrm{0},\mathrm{2}\pi\right]: \\ $$$${x}=\mathrm{0},\frac{\pi}{\mathrm{4}},\frac{\pi}{\mathrm{2}},\frac{\mathrm{3}\pi}{\mathrm{4}},\pi,\frac{\mathrm{5}\pi}{\mathrm{4}},\frac{\mathrm{3}\pi}{\mathrm{2}},\frac{\mathrm{7}\pi}{\mathrm{4}},\mathrm{2}\pi \\ $$
Commented by anonymous last updated on 13/Aug/20
$$\mathbb{GOD}\:\mathbb{BLESS}\:\mathbb{YOU}\: \\ $$