sin-5-x-2-sinx-1-x-0-2-

Question Number 75846 by behi83417@gmail.com last updated on 18/Dec/19

\sin^{5} x + \sqrt{2} sinx = 1, x \in [0, 2 π]

Answered by MJS last updated on 18/Dec/19

t = \sin x

t^{5} + \sqrt{2} t - 1 = 0

only approximation possible

only one real solution

t \approx . 634429

\Rightarrow x = . 687270 \lor x = 2.45432

Commented by behi83417@gmail.com last updated on 18/Dec/19

thank you proph : MJS .

this is my typo . the original question is :

\sin 5 x + \sqrt{2} sinx = 1 .

Answered by MJS last updated on 18/Dec/19

\sin 5 x + \sqrt{2} \sin x = 1

\sin 5 x = ({16 \sin}^{4} x - {20 \sin}^{2} x + 5) \sin x

\sin x = t

16 t^{5} - 20 t^{3} + (5 + \sqrt{2}) t = 1

t^{5} - \frac{5}{4} t^{3} + \frac{5 + \sqrt{2}}{16} t - \frac{1}{16} = 0

approximation

t_{1} \approx . 171177

t_{2} \approx . 551260

t_{3} \approx . 929365

t_{4, 5} \approx - . 825901 \pm . 174831 i

\Rightarrow

x_{11} \approx . 172024; x_{12} \approx 2.96957

x_{21} \approx . 583873; x_{22} \approx 2.55772

x_{31} \approx 1.19269; x_{32} \approx 1.94890

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