sin-4-x-sin-4-x-pi-4-1-4-x-0-2pi-

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

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