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Question Number 84134 by Rio Michael last updated on 09/Mar/20

find the general solution of the equation 2sin 3θ = sin 2θ

$$\mathrm{find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\mathrm{2sin}\:\mathrm{3}\theta\:=\:\mathrm{sin}\:\mathrm{2}\theta \\ $$

Answered by mr W last updated on 09/Mar/20

2 sin θ(3−4 sin^2 θ)=2 sin θ cos θ sin θ(4 cos^2 θ−cos θ−1)=0 ⇒sin θ=0 ⇒θ=kπ or ⇒4 cos^2 θ−cos θ−1=0 ⇒cos θ=((1±(√(17)))/8) ⇒θ=2kπ±cos^(−1) ((1+(√(17)))/8) ⇒θ=(2k+1)π±cos^(−1) (((√(17))−1)/8)

$$\mathrm{2}\:\mathrm{sin}\:\theta\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right)=\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta \\ $$$$\mathrm{sin}\:\theta\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{cos}\:\theta−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\theta={k}\pi \\ $$$${or} \\ $$$$\Rightarrow\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{cos}\:\theta−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{1}\pm\sqrt{\mathrm{17}}}{\mathrm{8}} \\ $$$$\Rightarrow\theta=\mathrm{2}{k}\pi\pm\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}+\sqrt{\mathrm{17}}}{\mathrm{8}} \\ $$$$\Rightarrow\theta=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\pm\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{17}}−\mathrm{1}}{\mathrm{8}} \\ $$

Commented by Rio Michael last updated on 09/Mar/20

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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