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Question Number 97994 by hardylanes last updated on 10/Jun/20
find all the values of θ, in the interval 0≤θ≤2π for which sin 3θ−sin θ=(√(3cos 2θ))
$${find}\:{all}\:{the}\:{values}\:{of}\:\theta,\:{in}\:{the}\:{interval}\:\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi \\ $$$${for}\:{which}\:\mathrm{sin}\:\mathrm{3}\theta−\mathrm{sin}\:\theta=\sqrt{\mathrm{3cos}\:\mathrm{2}\theta} \\ $$
Answered by Rio Michael last updated on 10/Jun/20
i pressume your question is sin 3θ − sin θ = (√3) cos 2θ for 0 ≤ θ ≤ 2π Recall sin A − sin B = 2 cos (((A + B)/2)) sin (((A−B)/2)) ⇒ 2 cos (((3θ + θ)/2)) sin (((3θ−θ)/2)) = (√3) cos 2θ ⇒ 2 cos 2θ sin θ = (√3) cos 2θ ⇒ 2cos 2θ sin θ−(√3) cos 2θ = 0 cos 2θ(2 sin θ−(√3) ) = 0 ⇔ cos 2θ = 0 ⇒ 2θ = 2πn ± (π/2) ⇒ θ = (1/2)(2πn ± (π/2)) n ∈ Z^+ sin θ = ((√3)/2) ⇒ πn + (−1)^n (π/3) n∈ Z^+ general solution θ = { (((1/2)(2πn ± (π/2)))),((πn +(−1)^n (π/3))) :} n ∈ Z^+ n = 0, ⇒θ = (π/4), (π/3) n = 1, ⇒ θ = ((4π)/3), ((3π)/4) ⇒ solution S = {(π/4),(π/3),((4π)/3),((3π)/4)}
$$\mathrm{i}\:\mathrm{pressume}\:\mathrm{your}\:\mathrm{question}\:\mathrm{is} \\ $$$$\:\mathrm{sin}\:\mathrm{3}\theta\:−\:\mathrm{sin}\:\theta\:=\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{for}\:\mathrm{0}\:\leqslant\:\theta\:\leqslant\:\mathrm{2}\pi \\ $$$$\mathcal{R}\mathrm{ecall}\:\mathrm{sin}\:{A}\:−\:\mathrm{sin}\:{B}\:=\:\mathrm{2}\:\mathrm{cos}\:\left(\frac{{A}\:+\:{B}}{\mathrm{2}}\right)\:\mathrm{sin}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right) \\ $$$$\:\Rightarrow\:\mathrm{2}\:\mathrm{cos}\:\left(\frac{\mathrm{3}\theta\:+\:\theta}{\mathrm{2}}\right)\:\mathrm{sin}\:\left(\frac{\mathrm{3}\theta−\theta}{\mathrm{2}}\right)\:=\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\:\Rightarrow\:\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{sin}\:\theta\:=\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\Rightarrow\:\mathrm{2cos}\:\mathrm{2}\theta\:\mathrm{sin}\:\theta−\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{2}\theta\:=\:\mathrm{0} \\ $$$$\:\:\mathrm{cos}\:\mathrm{2}\theta\left(\mathrm{2}\:\mathrm{sin}\:\theta−\sqrt{\mathrm{3}}\:\right)\:=\:\mathrm{0}\: \\ $$$$\Leftrightarrow\:\mathrm{cos}\:\mathrm{2}\theta\:=\:\mathrm{0}\:\Rightarrow\:\mathrm{2}\theta\:=\:\mathrm{2}\pi{n}\:\pm\:\frac{\pi}{\mathrm{2}}\:\Rightarrow\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\pi{n}\:\pm\:\frac{\pi}{\mathrm{2}}\right)\:{n}\:\in\:\mathbb{Z}^{+} \\ $$$$\:\:\:\:\mathrm{sin}\:\theta\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\Rightarrow\:\:\pi{n}\:+\:\left(−\mathrm{1}\right)^{{n}} \frac{\pi}{\mathrm{3}}\:{n}\in\:\mathbb{Z}^{+} \\ $$$$\mathrm{general}\:\mathrm{solution}\:\theta\:=\:\begin{cases}{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\pi{n}\:\pm\:\frac{\pi}{\mathrm{2}}\right)}\\{\pi{n}\:+\left(−\mathrm{1}\right)^{{n}} \frac{\pi}{\mathrm{3}}}\end{cases}\:{n}\:\in\:\mathbb{Z}^{+} \\ $$$${n}\:=\:\mathrm{0},\:\Rightarrow\theta\:=\:\frac{\pi}{\mathrm{4}},\:\frac{\pi}{\mathrm{3}} \\ $$$$\:{n}\:=\:\mathrm{1},\:\Rightarrow\:\theta\:=\:\frac{\mathrm{4}\pi}{\mathrm{3}},\:\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{solution}\:{S}\:=\:\left\{\frac{\pi}{\mathrm{4}},\frac{\pi}{\mathrm{3}},\frac{\mathrm{4}\pi}{\mathrm{3}},\frac{\mathrm{3}\pi}{\mathrm{4}}\right\} \\ $$
Commented by hardylanes last updated on 10/Jun/20
thanks☆
Commented by Rio Michael last updated on 10/Jun/20
you are welcome.
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}. \\ $$

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