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Question Number 106469 by ZiYangLee last updated on 05/Aug/20

$$\mathrm{Solve}{\cos }^2\mathrm{x}+{3\cos }^{2}2\mathrm{x}={\cos }^{2}3\mathrm{x}$$

Answered by mahdi last updated on 05/Aug/20

$$\cos \left(3\mathrm{x}\right)=\cos \left(2\mathrm{x}\right).\mathrm{cosx}-\sin \left(2\mathrm{x}\right).\mathrm{sinx}=$$$$({2\cos }^2\mathrm{x}-1).\mathrm{cosx}-(1-{\cos }^2\mathrm{x}).\mathrm{cosx}=$$$${3\cos }^3\mathrm{x}-2\mathrm{cosx}$$$$\cos \left(2\mathrm{x}\right)={2\cos }^2\mathrm{x}-1$$$${\cos }^2\mathrm{x}+{3\cos }^{2}2\mathrm{x}={\cos }^{2}3\mathrm{x}\Rightarrow$$$${\cos }^2\mathrm{x}+3{({2\cos }^2-1)}^2={({3\cos }^3\mathrm{x}-2\mathrm{cosx})}^2\Rightarrow$$$$\mathrm{get}\mathrm{cosx}=\mathrm{a}$$$${\mathrm{a}}^2+3{({2\mathrm{a}}^2-1)}^2={({3\mathrm{a}}^3-2\mathrm{a})}^2\Rightarrow$$$$0={3\mathrm{a}}^6-{8\mathrm{a}}^4+{5\mathrm{a}}^2-1$$$$$$

Commented by malwaan last updated on 06/Aug/20

$$b=a^2\Rightarrow 3b^3-8b^2+5b-1=0$$

Answered by john santu last updated on 05/Aug/20

$$3\cos {}^{2}2\mathrm{x}=\cos {}^{2}3\mathrm{x}-\cos {}^2\mathrm{x}$$$$3\cos {}^{2}2\mathrm{x}=(\cos 3\mathrm{x}+\cos \mathrm{x})(\cos 3\mathrm{x}-\cos \mathrm{x})$$$$3\cos {}^{2}2\mathrm{x}=2\cos 2\mathrm{xcos}\mathrm{x}(\cos 3\mathrm{x}-\cos \mathrm{x})$$$$\mathrm{case}\left(1\right)\cos 2\mathrm{x}=0,\mathrm{x}=\pm \frac{\pi }{4}+\mathrm{k}.180°$$$$\mathrm{case}\left(2\right)$$$$2\cos \mathrm{x}(\cos 3\mathrm{x}-\cos \mathrm{x})-3\cos 2\mathrm{x}=0$$$$2\cos \mathrm{x}(4\cos {}^3\mathrm{x}-4\cos \mathrm{x})-3(2\cos {}^2\mathrm{x}-1)=0$$$$\mathrm{let}\cos \mathrm{x}=\mathrm{t}$$$$\Rightarrow 2\mathrm{t}({4\mathrm{t}}^3-4\mathrm{t})-{6\mathrm{t}}^2+3=0$$$${8\mathrm{t}}^4-{14\mathrm{t}}^2+3=0;{\mathrm{t}}^2=\frac{14\pm 10}{16}$$$${\mathrm{t}}^2=\frac{3}{2}\left(\mathrm{rejected}\right);{\mathrm{t}}^2=\frac{1}{4}$$$$\mathrm{now}\mathrm{easy}\mathrm{to}\mathrm{solve}$$

Commented by ZiYangLee last updated on 05/Aug/20

$$\mathrm{thank}\mathrm{you}<33$$

Commented by john santu last updated on 05/Aug/20

$$⩾44.$$

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