solution-set-equation-sin-6-x-cos-6-x-1-4-sin-2-2x-

Question Number 140177 by liberty last updated on 05/May/21

solution set equation

\sin^{6} x + \cos^{6} x = \frac{1}{4} \sin^{2} 2 x

Answered by som(math1967) last updated on 05/May/21

{({s i n}^{2} x + {c o s}^{2} x)}^{3}

- 3 {s i n}^{2} {x c o s}^{2} x ({s i n}^{2} x + {c o s}^{2} x) = \frac{1}{4} {s i n}^{2} 2 x

1 - \frac{3}{4} {(2 s i n x c o s x)}^{2} = \frac{1}{4} {s i n}^{2} 2 x

1 = \frac{3}{4} {s i n}^{2} 2 x + \frac{1}{4} {s i n}^{2} 2 x

\Rightarrow {s i n}^{2} 2 x = 1

s i n 2 x = \pm 1

w h e n s i n 2 x = 1

x = (4 n + 1) \frac{π}{4}

s i n 2 x = - 1

x = (4 n - 1) \frac{π}{4}

Answered by liberty last updated on 05/May/21

\Rightarrow \sin^{6} x + \cos^{6} x = \frac{1}{4} . 4 \sin^{2} x \cos^{2} x

\Rightarrow (\sin^{2} x + \cos^{2} x) (\sin^{4} x - \sin^{2} xcos^{2} x + \cos^{4} x) = {(\sin x \cos x)}^{2}

\Rightarrow 1 - 3 \sin^{2} x \cos^{2} x = \sin^{2} x \cos^{2} x

\Rightarrow 4 \sin^{2} x \cos^{2} x = 1

\Rightarrow (\sin 2 x + 1) (\sin 2 x - 1) = 0

\Rightarrow {\begin{cases} \sin 2 x = - 1 \\ \sin 2 x = 1 \end{cases} \Rightarrow {\begin{cases} x = \frac{3 π}{4} + n π \\ x = \frac{π}{4} + n π \end{cases}

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