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

h ello solve it in] - π; π] and place solutions

in trigonometric circle .

\cos^{6} x + \sin^{6} x = \frac{3}{8} (\sqrt{3} \sin 4 x + \frac{8}{3})

please help me \dots

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 4 x

\frac{5}{8} + \frac{3}{8} \cos 4 x = 1 + \frac{3 \sqrt{3}}{8} \sin 4 x

\cos 4 x - \sqrt{3} \sin 4 x - 1 = 0

x = \frac{1}{2} \arctan t \Leftrightarrow t = \tan 2 x

- \frac{2 t (t + \sqrt{3})}{t^{2} + 1} = 0

\Rightarrow t_{1} = 0 \land t_{2} = - \sqrt{3}

\tan 2 x = 0 \Rightarrow x = - \frac{π}{2} \lor x = 0 \lor x = \frac{π}{2} \lor x = π

\tan 2 x = - \sqrt{3} \Rightarrow x = - \frac{2 π}{3} \lor x = - \frac{π}{6} \lor x = \frac{π}{3} \lor x = \frac{5 π}{6}

Commented by mathocean1 last updated on 22/Dec/19

please how have you done to have

the line number 5

\frac{5}{8} + \frac{3}{8} \cos 4 x

Commented by MJS last updated on 22/Dec/19

1 - {3 \sin}^{2} x \cos^{2} x = 1 - 3 {(\sin x c o s x)}^{2} =

[\sin x \cos x = \frac{1}{2} \sin 2 x

= 1 - 3 {(\frac{\sin 2 x}{2})}^{2} = 1 - \frac{3}{4} \sin^{2} 2 x =

[\sin^{2} x = \frac{1}{2} (1 - \cos 2 x)]

= 1 - \frac{3}{4} (\frac{1}{2} - \cos 4 x) = \frac{5}{8} + \frac{3}{8} \cos 4 x

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