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Question Number 160815 by cortano last updated on 07/Dec/21

$$\sec \left(3\mathrm{x}\right)-6\cos \left(3\mathrm{x}\right)=4\sin \left(3\mathrm{x}\right)$$$$\mathrm{find}\mathrm{the}\mathrm{solution}$$

Commented by blackmamba last updated on 07/Dec/21

$$\Rightarrow \frac{1}{\cos 3x}-6\cos 3x=4\sin 3x$$$$\Rightarrow \frac{1}{t}-6t=4\sqrt{1-t^2};[t=\cos 3x]$$$$\Rightarrow 1-6t^2=4t\sqrt{1-t^2}$$$$\Rightarrow 1-12t^2+36t^4=16t^2(1-t^2)$$$$\Rightarrow 52t^4-28t^2+1=0$$$$\Rightarrow t^2=\frac{28+\sqrt{{28}^2-4\times 52}}{104}=\frac{28+24}{104}$$$$\Rightarrow t^2=\frac{1}{2};t=\pm \frac{1}{2}\sqrt{2}$$$$case\left(1\right)t=\frac{1}{2}\sqrt{2}\Rightarrow \cos 3x=\cos 45°$$$$\Rightarrow x=\pm 15°+k.120°$$$$case\left(2\right)t=-\frac{1}{2}\sqrt{2}\Rightarrow \cos 3x=\cos 135°$$$$\Rightarrow x=\pm 45°+k.120°$$

Answered by mr W last updated on 08/Dec/21

$$\frac{1}{\cos 3x}-6\cos 3x=4\sin 3x$$$$1-6{\cos }^{2}3x=4\sin 3x\cos 3x$$$$1-3(\cos 6x+1)=2\sin 6x$$$$2\sin 6x+3\cos 6x=-2$$$$\frac{2}{\sqrt{13}}\sin 6x+\frac{3}{\sqrt{13}}\cos 6x=-\frac{2}{\sqrt{13}}$$$$with\alpha ={\tan }^{-1}\frac{3}{2}$$$$\sin (6x+\alpha )=-\cos \alpha =-\sin (\frac{\pi }{2}-\alpha )$$$$6x+\alpha =n\pi -{(-1)}^n(\frac{\pi }{2}-\alpha )$$$$\Rightarrow x=\frac{1}{6}\{n\pi -{(-1)}^n\frac{\pi }{2}+[{(-1)}^n-1]{\tan }^{-1}\frac{3}{2}\}$$$$=\{\begin{array}{l}\frac{(4k-1)\pi }{12}\\ \frac{(4k+3)\pi }{12}-\frac{1}{3}{\tan }^{-1}\frac{3}{2}\end{array}$$

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