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Question Number 97994 by hardylanes last updated on 10/Jun/20

$$findallthevaluesof\theta ,intheinterval0⩽\theta ⩽2\pi$$$$forwhich\sin 3\theta -\sin \theta =\sqrt{3\cos 2\theta }$$

Answered by Rio Michael last updated on 10/Jun/20

$$\mathrm{i}\mathrm{pressume}\mathrm{your}\mathrm{question}\mathrm{is}$$$$\sin 3\theta -\sin \theta =\sqrt{3}\cos 2\theta \mathrm{for}0⩽\theta ⩽2\pi$$$$\mathcal{R}\mathrm{ecall}\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$$$$\Rightarrow 2\cos \left(\frac{3\theta +\theta }{2}\right)\sin \left(\frac{3\theta -\theta }{2}\right)=\sqrt{3}\cos 2\theta$$$$\Rightarrow 2\cos 2\theta \sin \theta =\sqrt{3}\cos 2\theta$$$$\Rightarrow 2\cos 2\theta \sin \theta -\sqrt{3}\cos 2\theta =0$$$$\cos 2\theta (2\sin \theta -\sqrt{3})=0$$$$\iff \cos 2\theta =0\Rightarrow 2\theta =2\pi n\pm \frac{\pi }{2}\Rightarrow \theta =\frac{1}{2}(2\pi n\pm \frac{\pi }{2})n\in {\mathbb{Z}}^{+}$$$$\sin \theta =\frac{\sqrt{3}}{2}\Rightarrow \pi n+{(-1)}^n\frac{\pi }{3}n\in {\mathbb{Z}}^{+}$$$$\mathrm{general}\mathrm{solution}\theta =\{\begin{array}{l}\frac{1}{2}(2\pi n\pm \frac{\pi }{2})\\ \pi n+{(-1)}^n\frac{\pi }{3}\end{array}n\in {\mathbb{Z}}^{+}$$$$n=0,\Rightarrow \theta =\frac{\pi }{4},\frac{\pi }{3}$$$$n=1,\Rightarrow \theta =\frac{4\pi }{3},\frac{3\pi }{4}$$$$\Rightarrow \mathrm{solution}S=\{\frac{\pi }{4},\frac{\pi }{3},\frac{4\pi }{3},\frac{3\pi }{4}\}$$

Commented by hardylanes last updated on 10/Jun/20

thanks☆

Commented by Rio Michael last updated on 10/Jun/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}.$$

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