sin^4 x + sin^4 (x+(π/4)) = (1/4) x ∈ [ 0,2π ]


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Question Number 87488 by jagoll last updated on 04/Apr/20

$\sin^{4} x + \sin^{4} (x + \frac{π}{4}) = \frac{1}{4}$ $x \in [0, 2 π]$
Commented by john santu last updated on 05/Apr/20

${({2 \sin}^{2} x)}^{2} + {({2 \sin}^{2} (x + \frac{π}{4}))}^{2} = 1$ ${(1 - \cos 2 x)}^{2} + {(1 - \cos (2 x + \frac{π}{2}))}^{2} = 1$ ${(1 - \cos 2 x)}^{2} + {(1 + \sin 2 x)}^{2} = 1$ $1 - 2 \cos 2 x + \cos^{2} 2 x + 1 + 2 \sin 2 x + \sin^{2} 2 x = 1$ $2 \sin 2 x - 2 \cos 2 x = - 2$ $\cos 2 x - \sin 2 x = 1$ $\cos (2 x + \frac{π}{4}) = \frac{1}{\sqrt{2}} = \cos \frac{π}{4}$ $\Rightarrow 2 x = - \frac{π}{4} \pm \frac{π}{4} + 2 π n, n \in Z$ $\Rightarrow x = - \frac{π}{8} \pm \frac{π}{8} + π n$ $(+) \Rightarrow x = 0, π, 2 π$ $(-) \Rightarrow x = - \frac{π}{4} + π n; x = \frac{3 π}{4}, \frac{7 π}{4}$
	Answered by mind is power last updated on 04/Apr/20
	$sin(x+(π/4))=((√2)/2)(sin(x)+cos(x)) sin^2 (x+(π/4))=(1/2)(1+sin(2x)) sin^4 (x+(π/4))=(1/4)(1+sin^2 (2x)+2sin(2x)) ⇔sin^4 (x)+((2sin(2x)+sin^2 (2x))/4)=0 ⇔sin(x)(4sin^3 (x)+4cos(x)+4sin(x)cos^2 (x))=0 ⇔4sin(x)(sin(x)+cos(x))=0 ⇒4(√2)sin(x)sin(x+(π/4))=0 x∈{0,π,2π,((3π)/4),((7π)/4)}$
	$s i n (x + \frac{π}{4}) = \frac{\sqrt{2}}{2} (s i n (x) + c o s (x))$ ${s i n}^{2} (x + \frac{π}{4}) = \frac{1}{2} (1 + s i n (2 x))$ ${s i n}^{4} (x + \frac{π}{4}) = \frac{1}{4} (1 + {s i n}^{2} (2 x) + 2 s i n (2 x))$ $\Leftrightarrow {s i n}^{4} (x) + \frac{2 s i n (2 x) + {s i n}^{2} (2 x)}{4} = 0$ $\Leftrightarrow s i n (x) (4 {s i n}^{3} (x) + 4 c o s (x) + 4 s i n (x) {c o s}^{2} (x)) = 0$ $\Leftrightarrow 4 s i n (x) (s i n (x) + c o s (x)) = 0$ $\Rightarrow 4 \sqrt{2} s i n (x) s i n (x + \frac{π}{4}) = 0$ $x \in {0, π, 2 π, \frac{3 π}{4}, \frac{7 π}{4}}$
	Answered by TANMAY PANACEA. last updated on 04/Apr/20

	${(\frac{1 - c o s 2 x}{2})}^{2} + {(\frac{1 - c o s (\frac{π}{2} + 2 x)}{2})}^{2} = \frac{1}{4}$ $\frac{1 - 2 c o s 2 x + {c o s}^{2} 2 x}{4} + {(\frac{1 + s i n 2 x}{2})}^{2} = \frac{1}{4}$ $1 - 2 c o s 2 x + {c o s}^{2} 2 x + 1 + 2 s i n 2 x + {s i n}^{2} 2 x = 1$ $3 - 2 c o s 2 x + 2 s i n 2 x = 1$ $1 - c o s 2 x + s i n 2 x = 0$ $1 = {(c o s 2 x - s i n 2 x)}^{2}$ $1 = 1 - s i n 4 x$ $s i n 4 x = 0 = s i n 0 \to x = 0$
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