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Question Number 144237 by mathdanisur last updated on 23/Jun/21

Answered by MJS_new last updated on 23/Jun/21

$$t=\tan \frac{x}{2}$$$$3\times \frac{2t}{t^2+1}-4\times \frac{t^2-1}{t^2+1}=3$$$$t^2-\frac{6}{7}t-\frac{1}{7}=0$$$$t=-\frac{1}{7}\vee t=1$$$$x=2n\pi -2\arctan \frac{1}{7}\vee x=2n\pi +\frac{\pi }{2}$$

Commented by mathdanisur last updated on 23/Jun/21

$$ThankyouSir$$

Answered by mathmax by abdo last updated on 23/Jun/21

$$3\mathrm{sinx}+4\mathrm{cosx}=\sqrt{3^2+4^2}(\frac{3}{\sqrt{(...)}}\mathrm{sinx}+\frac{4}{\sqrt{(...)}}\mathrm{cosx})$$$$=5(\mathrm{cosx}\cos \alpha +\mathrm{sinx}\sin \alpha )\mathrm{with}\cos \alpha =\frac{4}{5}\mathrm{and}\sin \alpha =\frac{3}{5}\Rightarrow$$$$\tan \alpha =\frac{3}{4}\Rightarrow \alpha =\arctan \left(\frac{3}{4}\right)$$$$\mathrm{e}\Rightarrow 5\cos (\mathrm{x}-\alpha )=3\Rightarrow \cos (\mathrm{x}-\alpha )=\frac{3}{5}\mathrm{let}\cos \lambda =\frac{3}{5}$$$$\mathrm{e}\Rightarrow \mathrm{x}-\alpha =\lambda +2\mathrm{k}\pi \mathrm{or}\mathrm{x}-\alpha =-\lambda +2\mathrm{k}\pi \Rightarrow$$$$\mathrm{x}=\alpha \stackrel{-}{+}\lambda +2\mathrm{k}\pi \Rightarrow \mathrm{x}=\arctan \left(\frac{3}{4}\right)\stackrel{-}{+}\mathrm{arcos}\left(\frac{3}{5}\right)+2\mathrm{k}\pi /(\mathrm{k}\in \mathrm{Z})$$

Commented by mathdanisur last updated on 23/Jun/21

$$ThankyouSir$$

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