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Question Number 85489 by oustmuchiya@gmail.com last updated on 22/Mar/20

Find all angles between 0° and 360°, for which 8sinθ=3cos^2 θ

$${Find}\:{all}\:{angles}\:{between}\:\mathrm{0}°\:{and}\:\mathrm{360}°,\:{for}\:{which}\:\mathrm{8}{sin}\theta=\mathrm{3}{cos}^{\mathrm{2}} \theta \\ $$

Commented by Tony Lin last updated on 22/Mar/20

8sinθ=3(1−sin^2 θ)  3sin^2 θ+8sinθ−3=0  (3sinθ−1)(sinθ+3)=0  sinθ=(1/3)  θ ≒ 19.47°&160.53°

$$\mathrm{8}{sin}\theta=\mathrm{3}\left(\mathrm{1}−{sin}^{\mathrm{2}} \theta\right) \\ $$$$\mathrm{3}{sin}^{\mathrm{2}} \theta+\mathrm{8}{sin}\theta−\mathrm{3}=\mathrm{0} \\ $$$$\left(\mathrm{3}{sin}\theta−\mathrm{1}\right)\left({sin}\theta+\mathrm{3}\right)=\mathrm{0} \\ $$$${sin}\theta=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\theta\:\fallingdotseq\:\mathrm{19}.\mathrm{47}°\&\mathrm{160}.\mathrm{53}° \\ $$

Commented by john santu last updated on 22/Mar/20

8sin θ = 3(1−sin^2 θ)  3sin^2 θ + 8sin θ−3=0  (3sin θ−1)(sin θ+3)=0  sin θ = (1/3)⇒θ=  { ((sin^(−1) ((1/3))+2kπ)),((π−sin^(−1) ((1/3))+2kπ)) :}

$$\mathrm{8sin}\:\theta\:=\:\mathrm{3}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta\right) \\ $$$$\mathrm{3sin}\:^{\mathrm{2}} \theta\:+\:\mathrm{8sin}\:\theta−\mathrm{3}=\mathrm{0} \\ $$$$\left(\mathrm{3sin}\:\theta−\mathrm{1}\right)\left(\mathrm{sin}\:\theta+\mathrm{3}\right)=\mathrm{0} \\ $$$$\mathrm{sin}\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow\theta=\:\begin{cases}{\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{2}{k}\pi}\\{\pi−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{2}{k}\pi}\end{cases} \\ $$

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