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Question Number 75976 by mathocean1 last updated on 21/Dec/19

$$\mathrm{h}\mathrm{ello}\mathrm{solve}\mathrm{it}\mathrm{in}]-\pi ;\pi ]\mathrm{and}\mathrm{place}\mathrm{solutions}$$$$\mathrm{in}\mathrm{trigonometric}\mathrm{circle}.$$$${\cos }^6\mathrm{x}+{\sin }^6\mathrm{x}=\frac{3}{8}(\sqrt{3}\sin 4\mathrm{x}+\frac{8}{3})$$$$\mathrm{please}\mathrm{help}\mathrm{me}...$$

Commented by MJS last updated on 21/Dec/19

$${\cos }^{6}x+{\sin }^{6}x=$$$$[{\cos }^{2}x=1-{\sin }^{2}x]$$$$={3\sin }^{4}x-{3\sin }^{2}x+1=1+{3\sin }^{2}x({\sin }^{2}x-1)=$$$$=1-{3\sin }^{2}x(1-{\sin }^{2}x)=1-{3\sin }^{2}x{\cos }^{2}x=$$$$=\frac{5}{8}+\frac{3}{8}\cos 4x$$$$$$$$\frac{5}{8}+\frac{3}{8}\cos 4x=1+\frac{3\sqrt{3}}{8}\sin 4x$$$$\cos 4x-\sqrt{3}\sin 4x-1=0$$$$x=\frac{1}{2}\arctan t\iff t=\tan 2x$$$$-\frac{2t(t+\sqrt{3})}{t^2+1}=0$$$$\Rightarrow t_1=0\wedge t_2=-\sqrt{3}$$$$\tan 2x=0\Rightarrow x=-\frac{\pi }{2}\vee x=0\vee x=\frac{\pi }{2}\vee x=\pi$$$$\tan 2x=-\sqrt{3}\Rightarrow x=-\frac{2\pi }{3}\vee x=-\frac{\pi }{6}\vee x=\frac{\pi }{3}\vee x=\frac{5\pi }{6}$$

Commented by mathocean1 last updated on 22/Dec/19

$$\mathrm{please}\mathrm{how}\mathrm{have}\mathrm{you}\mathrm{done}\mathrm{to}\mathrm{have}$$$$\mathrm{the}\mathrm{line}\mathrm{number}5$$$$\frac{5}{8}+\frac{3}{8}\cos 4x$$

Commented by MJS last updated on 22/Dec/19

$$1-{3\sin }^{2}x{\cos }^{2}x=1-3{\left(\sin xcosx\right)}^2=$$$$[\sin x\cos x=\frac{1}{2}\sin 2x$$$$=1-3{\left(\frac{\sin 2x}{2}\right)}^2=1-\frac{3}{4}{\sin }^{2}2x=$$$$[{\sin }^{2}x=\frac{1}{2}(1-\cos 2x)]$$$$=1-\frac{3}{4}(\frac{1}{2}-\cos 4x)=\frac{5}{8}+\frac{3}{8}\cos 4x$$

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