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

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