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Question Number 132347 by Study last updated on 13/Feb/21

$$sin(2x+\frac{\pi }{4})-sin(2x+\frac{\pi }{3})=0x=?$$

Answered by EDWIN88 last updated on 13/Feb/21

$$\sin \mathrm{X}-\sin \mathrm{Y}=2\cos \left(\frac{\mathrm{X}+\mathrm{Y}}{2}\right)\sin \left(\frac{\mathrm{X}-\mathrm{Y}}{2}\right)$$$$\mathrm{where}\{\begin{array}{l}\mathrm{X}=2\mathrm{x}+\frac{\pi }{4}\\ \mathrm{Y}=2\mathrm{x}+\frac{\pi }{3}\end{array}$$$$\sin (2\mathrm{x}+\frac{\pi }{4})-\sin (2\mathrm{x}+\frac{\pi }{3})=0$$$$2\cos (2\mathrm{x}+\frac{7\pi }{24}).\sin (-\frac{\pi }{24})=0$$$$\cos (2\mathrm{x}+\frac{7\pi }{24})=0=\cos \frac{\pi }{2}$$$$2\mathrm{x}=\pm \frac{\pi }{2}-\frac{7\pi }{24}+2\mathrm{n}\pi$$$$\mathrm{x}=\pm \frac{\pi }{4}-\frac{7\pi }{48}+\mathrm{n}\pi \Rightarrow \mathrm{x}=\{\begin{array}{l}\frac{5\pi }{48}+\mathrm{n}\pi \\ -\frac{19}{48}+\mathrm{n}\pi \end{array}$$

Answered by Ar Brandon last updated on 13/Feb/21

$$\sin (2\mathrm{x}+\frac{\pi }{4})-\sin (2\mathrm{x}+\frac{\pi }{3})=0\Rightarrow \sin (2\mathrm{x}+\frac{\pi }{4})=\sin (2\mathrm{x}+\frac{\pi }{3})$$$$\{\begin{array}{l}2\mathrm{x}+\frac{\pi }{4}=2\mathrm{x}+\frac{\pi }{3}\\ 2\mathrm{x}+\frac{\pi }{4}=\pi -(2\mathrm{x}+\frac{\pi }{3})\end{array}\Rightarrow 4\mathrm{x}=\pi -\frac{\pi }{4}-\frac{\pi }{3}=\frac{5\pi }{12}\left[2\pi \right]$$$$\Rightarrow \mathrm{x}=\frac{5\pi }{48}\left[\frac{\pi }{2}\right]$$

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