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Find-the-values-of-x-in-the-range-0-to-360-for-which-sin-3x-sin-x-2cos-2x-1-

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Question Number 13903 by tawa tawa last updated on 24/May/17
Find the values of x in the range 0° to 360° for which sin(3x)sin(x) = 2cos(2x) + 1
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{range}\:\mathrm{0}°\:\mathrm{to}\:\mathrm{360}°\:\mathrm{for}\:\mathrm{which}\: \\ $$$$\mathrm{sin}\left(\mathrm{3x}\right)\mathrm{sin}\left(\mathrm{x}\right)\:=\:\mathrm{2cos}\left(\mathrm{2x}\right)\:+\:\mathrm{1} \\ $$
Commented by myintkhaing last updated on 26/May/17
Please in the range 0° to 360° means 0°<x<360° or 0°≤x≤360° ??
$${Please}\:{in}\:{the}\:{range}\:\mathrm{0}°\:{to}\:\mathrm{360}°\:{means} \\ $$$$\mathrm{0}°<{x}<\mathrm{360}°\:{or}\:\mathrm{0}°\leqslant{x}\leqslant\mathrm{360}°\:?? \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17
(3sinx−4sin^3 x)sinx=2(1−2sin^2 x)+1 sinx=t⇒3t^2 −4t^4 =−4t^2 +3⇒ 4t^4 −7t^2 +3=0⇒t^2 =1,(3/4)⇒t=sinx=±1,±((√3)/2) sinx=1⇒x=2kπ+(π/2) sinx=−1⇒x=2kπ+((3π)/2) sinx=((√3)/2)⇒x=2kπ+(π/3),2kπ+((2π)/3) sinx=−((√3)/2)⇒x=2kπ+((4π)/3),2kπ+((5π)/3) ⇒x=(π/3),(π/2),((2π)/3),((4π)/3),((3π)/2),((5π)/3)∈(0,2π) .■
$$\left(\mathrm{3}{sinx}−\mathrm{4}{sin}^{\mathrm{3}} {x}\right){sinx}=\mathrm{2}\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} {x}\right)+\mathrm{1} \\ $$$${sinx}={t}\Rightarrow\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}{t}^{\mathrm{4}} =−\mathrm{4}{t}^{\mathrm{2}} +\mathrm{3}\Rightarrow \\ $$$$\mathrm{4}{t}^{\mathrm{4}} −\mathrm{7}{t}^{\mathrm{2}} +\mathrm{3}=\mathrm{0}\Rightarrow{t}^{\mathrm{2}} =\mathrm{1},\frac{\mathrm{3}}{\mathrm{4}}\Rightarrow{t}={sinx}=\pm\mathrm{1},\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${sinx}=\mathrm{1}\Rightarrow{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}} \\ $$$${sinx}=−\mathrm{1}\Rightarrow{x}=\mathrm{2}{k}\pi+\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$${sinx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{3}},\mathrm{2}{k}\pi+\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$${sinx}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow{x}=\mathrm{2}{k}\pi+\frac{\mathrm{4}\pi}{\mathrm{3}},\mathrm{2}{k}\pi+\frac{\mathrm{5}\pi}{\mathrm{3}} \\ $$$$\Rightarrow{x}=\frac{\pi}{\mathrm{3}},\frac{\pi}{\mathrm{2}},\frac{\mathrm{2}\pi}{\mathrm{3}},\frac{\mathrm{4}\pi}{\mathrm{3}},\frac{\mathrm{3}\pi}{\mathrm{2}},\frac{\mathrm{5}\pi}{\mathrm{3}}\in\left(\mathrm{0},\mathrm{2}\pi\right)\:\:\:\:.\blacksquare \\ $$
Commented by tawa tawa last updated on 25/May/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by ajfour last updated on 25/May/17
i think x=((2π)/3), ((5π)/3) also qualify in the given range .
$${i}\:{think}\:\:\:{x}=\frac{\mathrm{2}\pi}{\mathrm{3}},\:\frac{\mathrm{5}\pi}{\mathrm{3}}\:{also}\:{qualify}\: \\ $$$${in}\:{the}\:{given}\:{range}\:. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17
you are right sir.thanks.
$${you}\:{are}\:{right}\:{sir}.{thanks}. \\ $$
Answered by ajfour last updated on 25/May/17
2sin (3x)sin (x)=4cos (2x)+2 cos (2x)−cos (4x)=4cos (2x)+2 cos (4x)+3cos (2x)+2=0 2cos^2 (2x)−1+3cos (2x)+2=0 let t=cos (2x), then 2t^2 +3t+1=0 (2t+1)(t+1)=0 t=−(1/2) ; t=−1 ⇒ cos (2x)=−(1/2) where 0≤2x≤4π 2x=π±(π/3), 3π±(π/3) x=(π/3), ((2π)/3), ((4π)/3), ((5π)/3) and if cos (2x)=−1 2x=π, 3π x=(π/2), ((3π)/2) hence if 0≤x≤2π , then x=(𝛑/3), (𝛑/2), ((2𝛑)/3), ((4𝛑)/3), ((3𝛑)/2), ((5𝛑)/3) .
$$\mathrm{2sin}\:\left(\mathrm{3}{x}\right)\mathrm{sin}\:\left({x}\right)=\mathrm{4cos}\:\left(\mathrm{2}{x}\right)+\mathrm{2} \\ $$$$\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\mathrm{cos}\:\left(\mathrm{4}{x}\right)=\mathrm{4cos}\:\left(\mathrm{2}{x}\right)+\mathrm{2} \\ $$$$\mathrm{cos}\:\left(\mathrm{4}{x}\right)+\mathrm{3cos}\:\left(\mathrm{2}{x}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{2cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)−\mathrm{1}+\mathrm{3cos}\:\left(\mathrm{2}{x}\right)+\mathrm{2}=\mathrm{0} \\ $$$${let}\:{t}=\mathrm{cos}\:\left(\mathrm{2}{x}\right),\:{then} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{2}{t}+\mathrm{1}\right)\left({t}+\mathrm{1}\right)=\mathrm{0} \\ $$$${t}=−\frac{\mathrm{1}}{\mathrm{2}}\:\:;\:\:\:{t}=−\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{cos}\:\left(\mathrm{2}{x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\:\:{where}\:\mathrm{0}\leqslant\mathrm{2}{x}\leqslant\mathrm{4}\pi \\ $$$$\mathrm{2}{x}=\pi\pm\frac{\pi}{\mathrm{3}},\:\mathrm{3}\pi\pm\frac{\pi}{\mathrm{3}} \\ $$$${x}=\frac{\pi}{\mathrm{3}},\:\frac{\mathrm{2}\pi}{\mathrm{3}},\:\frac{\mathrm{4}\pi}{\mathrm{3}},\:\frac{\mathrm{5}\pi}{\mathrm{3}} \\ $$$${and}\:{if}\:\:\:\mathrm{cos}\:\left(\mathrm{2}{x}\right)=−\mathrm{1} \\ $$$$\mathrm{2}{x}=\pi,\:\mathrm{3}\pi \\ $$$${x}=\frac{\pi}{\mathrm{2}},\:\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$${hence}\:\:{if}\:\:\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\pi\:,\:{then} \\ $$$$\:\:\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{3}},\:\frac{\boldsymbol{\pi}}{\mathrm{2}},\:\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{3}},\:\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{3}},\:\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{2}},\:\frac{\mathrm{5}\boldsymbol{\pi}}{\mathrm{3}}\:\:. \\ $$
Commented by mrW1 last updated on 25/May/17
yes, there are 6 solutions in the range (0, 2π)
$${yes},\:{there}\:{are}\:\mathrm{6}\:{solutions}\:{in}\:{the} \\ $$$${range}\:\left(\mathrm{0},\:\mathrm{2}\pi\right) \\ $$
Commented by ajfour last updated on 25/May/17
thanks.
$${thanks}.\: \\ $$

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