solution set equation sin^6 x + cos^6 x = (1/4)sin^2 2x


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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
	$⇒sin^6 x+cos^6 x=(1/4).4sin^2 x cos^2 x ⇒(sin^2 x+cos^2 x)(sin^4 x−sin^2 xcos^2 x+cos^4 x)=(sin x cos x)^2 ⇒1−3sin^2 x cos^2 x = sin^2 x cos^2 x ⇒4sin^2 x cos^2 x = 1 ⇒(sin 2x+1)(sin 2x−1)=0 ⇒ { ((sin 2x=−1)),((sin 2x=1)) :} ⇒ { ((x=((3π)/4)+nπ)),((x=(π/4)+nπ)) :}$
	$\Rightarrow \sin^{6} x + \cos^{6} x = \frac{1}{4} . 4 \sin^{2} x \cos^{2} x$ $\Rightarrow (\sin^{2} x + \cos^{2} x) (\sin^{4} x - \sin^{2} xcos^{2} x + \cos^{4} x) = {(\sin x \cos x)}^{2}$ $\Rightarrow 1 - 3 \sin^{2} x \cos^{2} x = \sin^{2} x \cos^{2} x$ $\Rightarrow 4 \sin^{2} x \cos^{2} x = 1$ $\Rightarrow (\sin 2 x + 1) (\sin 2 x - 1) = 0$ $\Rightarrow {\begin{cases} \sin 2 x = - 1 \\ \sin 2 x = 1 \end{cases} \Rightarrow {\begin{cases} x = \frac{3 π}{4} + n π \\ x = \frac{π}{4} + n π \end{cases}$
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