Question Number 81057 by mathocean1 last updated on 09/Feb/20
![solve in [−π;π] (E): sin3x=−sin2x](Q81057.png)
$${solve}\:\mathrm{in}\:\left[−\pi;\pi\right]\: \\ $$$$\left({E}\right):\:{sin}\mathrm{3}{x}=−{sin}\mathrm{2}{x} \\ $$
Commented by jagoll last updated on 09/Feb/20
$$\mathrm{sin}\:\mathrm{3}{x}+\mathrm{sin}\:\mathrm{2}{x}\:=\mathrm{0} \\ $$$$\mathrm{2sin}\:\left(\frac{\mathrm{5}{x}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{sin}\:\left(\frac{\mathrm{5}{x}}{\mathrm{2}}\right)=\mathrm{0}\:=\mathrm{sin}\:\mathrm{0} \\ $$$$\frac{\mathrm{5}{x}}{\mathrm{2}}\:=\:\mathrm{2}{n}\pi\:\Rightarrow\:{x}\:=\:\frac{\mathrm{4}{n}\pi}{\mathrm{5}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)=\mathrm{0}\:\Rightarrow\frac{{x}}{\mathrm{2}}\:=\pm\frac{\pi}{\mathrm{2}}+\mathrm{2}{n}\pi \\ $$$${x}\:=\:\pm\:\pi+\mathrm{4}{n}\pi \\ $$
Commented by mr W last updated on 09/Feb/20
![if sin α=−sin β, it means α=(2k+1)π+β or α=2kπ−β sin3x=−sin2x 3x=(2k+1)π+2x or 3x=2kπ−2x x=(2k+1)π or x=(2/5)kπ within [−π,π] we have x=−π,π,−((4π)/5),((2π)/5),0,((2π)/5),((4π)/5)](Q81063.png)
$${if}\:\mathrm{sin}\:\alpha=−\mathrm{sin}\:\beta,\:{it}\:{means} \\ $$$$\alpha=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi+\beta\:{or}\:\alpha=\mathrm{2}{k}\pi−\beta \\ $$$$ \\ $$$${sin}\mathrm{3}{x}=−{sin}\mathrm{2}{x} \\ $$$$\mathrm{3}{x}=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi+\mathrm{2}{x}\:{or}\:\mathrm{3}{x}=\mathrm{2}{k}\pi−\mathrm{2}{x} \\ $$$${x}=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\:{or}\:{x}=\frac{\mathrm{2}}{\mathrm{5}}{k}\pi \\ $$$${within}\:\left[−\pi,\pi\right]\:{we}\:{have} \\ $$$${x}=−\pi,\pi,−\frac{\mathrm{4}\pi}{\mathrm{5}},\frac{\mathrm{2}\pi}{\mathrm{5}},\mathrm{0},\frac{\mathrm{2}\pi}{\mathrm{5}},\frac{\mathrm{4}\pi}{\mathrm{5}} \\ $$
Commented by jagoll last updated on 09/Feb/20
$${if}\:\mathrm{sin}\:\alpha\:=\:\mathrm{cos}\:\beta\:? \\ $$
Commented by mr W last updated on 09/Feb/20
$$\alpha+\beta=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}\:{or}\:\beta−\alpha=\mathrm{2}{k}\pi−\frac{\pi}{\mathrm{2}} \\ $$$${i}.{e}.\:\beta=\mathrm{2}{k}\pi\pm\left(\frac{\pi}{\mathrm{2}}−\alpha\right) \\ $$$${you}\:{can}\:{see}\:{all}\:{these}\:{on}\:{the}\:{graph}\:{of} \\ $$$${the}\:{functions}. \\ $$
Commented by jagoll last updated on 09/Feb/20
$${the}\:{same}\:{case}\:{for}\:\mathrm{sin}\:\alpha\:=−\mathrm{cos}\:\beta\:{sir} \\ $$
Commented by mr W last updated on 09/Feb/20
$${you}\:{can}\:{figure}\:{out}\:{all}\:{cases}\:{by}\:{yourself} \\ $$$${according}\:{to}\:{the}\:{graph}\:{of}\:{the}\:{functions}. \\ $$$$ \\ $$$$\mathrm{sin}\:\alpha\:=−\mathrm{cos}\:\beta\:\Rightarrow\mathrm{sin}\:\left(−\alpha\right)=\mathrm{cos}\:\beta \\ $$$${using}\:{that}\:{what}\:{we}\:{already}\:{know}\:{above}, \\ $$$${we}\:{get} \\ $$$$\Rightarrow\beta=\mathrm{2}{k}\pi\pm\left(\frac{\pi}{\mathrm{2}}+\alpha\right) \\ $$
Commented by john santu last updated on 09/Feb/20
$$\mathrm{sin}\:\mathrm{3}{x}\:=\:\mathrm{sin}\:\left(−\mathrm{2}{x}\right)\: \\ $$$$\mathrm{3}{x}\:=\:−\mathrm{2}{x}+\:\mathrm{2}{k}\pi\:\: \\ $$$$\mathrm{5}{x}\:=\:\mathrm{2}{k}\pi\:\Rightarrow{x}\:=\:\frac{\mathrm{2}{k}\pi}{\mathrm{5}} \\ $$$${or}\:\mathrm{3}{x}\:=\:\pi+\mathrm{2}{x}+\mathrm{2}{k}\pi \\ $$$${x}\:=\:\pi+\mathrm{2}{k}\pi\: \\ $$$$ \\ $$