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Question Number 41895 by lucha116 last updated on 15/Aug/18

$$\sqrt{3}sin3x-cos3x+2sin\frac{9x}{4}=4$$

Commented by MJS last updated on 15/Aug/18

$$\mathrm{the}\mathrm{answer}\mathrm{to}\mathrm{this}\mathrm{question}\mathrm{is}\mathrm{included}\mathrm{in}$$$$\mathrm{the}\mathrm{former}\mathrm{similar}\mathrm{one}.\mathrm{there}\mathrm{the}\mathrm{maximum}$$$$\mathrm{was}\left(\begin{array}{c}\frac{2\pi }{9}+\frac{8\pi }{3}z\\ 2\end{array}\right)\mathrm{with}z\in \mathbb{Z},\mathrm{here}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{zeros}$$$$\Rightarrow x=\frac{2\pi }{9}+\frac{8\pi }{3}z$$

Answered by MrW3 last updated on 15/Aug/18

$$2(\frac{\sqrt{3}}{2}sin3x-\frac{1}{2}cos3x)+2sin\frac{9x}{4}=4$$$$2(\cos \frac{\pi }{6}sin3x-\sin \frac{\pi }{6}cos3x)+2sin\frac{9x}{4}=4$$$$2sin(3x-\frac{\pi }{6})+2sin\frac{9x}{4}=4$$$$sin(3x-\frac{\pi }{6})+sin\frac{9x}{4}=2$$$$\left(1\right)$$$$sin(3x-\frac{\pi }{6})=1$$$$\Rightarrow 3x-\frac{\pi }{6}=2n\pi +\frac{\pi }{2}$$$$\Rightarrow x=\frac{2}{3}n\pi +\frac{2\pi }{9}...\left(i\right)$$$$\left(2\right)$$$$sin\frac{9x}{4}=1$$$$\Rightarrow \frac{9x}{4}=2m\pi +\frac{\pi }{2}$$$$\Rightarrow x=\frac{8}{9}m\pi +\frac{2\pi }{9}...\left(ii\right)$$$$sothat\left(i\right)and\left(ii\right)arethesame,$$$$\frac{2}{3}n=\frac{8}{9}m$$$$n=\frac{4}{3}m$$$$\Rightarrow m=3k$$$$\Rightarrow n=4k$$$$$$$$solutionis:$$$$x=\frac{8k}{3}\pi +\frac{2\pi }{9}=\frac{(24k+2)\pi }{9}withk=anyinteger$$

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