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Question Number 41878 by lucha116 last updated on 14/Aug/18

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

Answered by MJS last updated on 15/Aug/18

$$f\left(x\right)=\sqrt{3}\sin 3x-\cos 3x+2\sin \frac{9x}{4}-2$$$$\mathrm{found}\mathrm{by}\mathrm{approximation}:$$$$\max \left(f\left(x\right)\right)=\left(\begin{array}{c}\frac{2\pi }{9}+\frac{8\pi }{3}n\\ 2\end{array}\right)\mathrm{with}n\in \mathbb{Z}$$$$\Rightarrow \mathrm{period}\mathrm{length}=\frac{8\pi }{3}$$$$\mathrm{shift}\frac{2\pi }{9}\mathrm{to}\mathrm{the}\mathrm{left}\mathrm{to}\mathrm{get}\mathrm{it}\mathrm{symmetric}\mathrm{to}x=0$$$$x=r+\frac{2\pi }{9}$$$$f\left(r\right)=\sqrt{3}\sin (3r+\frac{2\pi }{3})-\cos (3r+\frac{2\pi }{3})+2\sin (\frac{9r}{4}+\frac{\pi }{2})-2$$$$f\left(r\right)=2(\cos 3r+\cos \frac{9r}{4}-1)$$$$s=\frac{4r}{3}$$$$f\left(s\right)=2(\cos 4s+\cos 3s-1)$$$$\cos 4s+\cos 3s-1=0$$$$s=\frac{\pi }{2}+\pi z\Rightarrow x=\frac{8\pi }{9}+\frac{4\pi }{3}z$$$$\mathrm{approximately}$$$$s\approx .297356+2\pi z\Rightarrow x\approx 1.09461+\frac{8\pi }{3}z$$$$s\approx 1.93560+2\pi z\Rightarrow x\approx 3.27894+\frac{8\pi }{3}z$$$$(z\in \mathbb{Z})$$$$\mathrm{exactly}$$$$\cos 4s+\cos 3s-1=0$$$$\mathrm{s}=2\arctan t$$$$\Rightarrow (t-1)(t+1)(t^6+19t^4-45t^2+1)=0$$$$t=\pm 1\Rightarrow s=\pm \frac{\pi }{2}$$$$t^6+19t^4-45t^2+1=0$$$$\mathrm{this}\mathrm{is}\mathrm{symm}.\mathrm{to}x=0,\mathrm{we}\mathrm{only}\mathrm{need}\mathrm{positive}$$$$\mathrm{solutions}$$$$t=\sqrt{u}$$$$u^3+19u^2-45u+1=0$$$$u=v-\frac{19}{3}$$$$v^3-\frac{496}{3}v+\frac{21440}{27}=0$$$$\mathrm{trigonometric}\mathrm{method}\mathrm{gives}$$$$v_1=\frac{8\sqrt{31}}{3}\sin \left(\frac{\arcsin \left(\frac{335\sqrt{31}}{1922}\right)}{3}\right)$$$$v_2=\frac{8\sqrt{31}}{3}\cos (\frac{\arcsin \left(\frac{335\sqrt{31}}{1922}\right)}{3}+\frac{\pi }{6})$$$$v_3=-\frac{8\sqrt{31}}{3}\sin (\frac{\arcsin \left(\frac{335\sqrt{31}}{1922}\right)}{3}+\frac{\pi }{3})<0\Rightarrow \mathrm{not}\mathrm{useable}$$$$\mathrm{these}\mathrm{cannot}\mathrm{be}\mathrm{further}\mathrm{simplfied},\mathrm{so}\mathrm{we}\mathrm{again}$$$$\mathrm{have}\mathrm{to}\mathrm{approximate}$$$$\mathrm{going}\mathrm{back}\mathrm{all}\mathrm{the}\mathrm{way}\mathrm{gives}\mathrm{the}\mathrm{same}\mathrm{solutiond}$$$$\mathrm{as}\mathrm{above}$$

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