Solve-the-equation-sinx-3-sinx-cosx-

Question Number 155244 by mathdanisur last updated on 27/Sep/21

Solve the equation :

{(\sin x)}^{3} + \sin x = \cos x

Answered by ajfour last updated on 27/Sep/21

\tan x (\frac{1}{1 + \frac{1}{\tan^{2} x}} + 1) = 1

m (2 m^{2} + 1) = 1 + m^{2}

2 m^{3} - m^{2} + m - 1 = 0

x = \tan^{- 1} m + k π

Commented by mathdanisur last updated on 27/Sep/21

very nice S er, thank you

Commented by mathdanisur last updated on 27/Sep/21

Thanks S er, but m = ?

Commented by mr W last updated on 27/Sep/21

m^{3} - \frac{m^{2}}{2} + \frac{m}{2} - \frac{1}{2} = 0

m = n + \frac{1}{6}

n^{3} + \frac{5 n}{12} - \frac{23}{54} = 0

m = \tan x = \frac{1}{6} + \sqrt[3]{\frac{\sqrt{249}}{72} + \frac{23}{108}} - \sqrt[3]{\frac{\sqrt{249}}{72} - \frac{23}{108}} \approx 0.739

x = k π + \tan^{- 1} (\frac{1}{6} + \sqrt[3]{\frac{\sqrt{249}}{72} + \frac{23}{108}} - \sqrt[3]{\frac{\sqrt{249}}{72} - \frac{23}{108}})

Answered by MJS_new last updated on 27/Sep/21

\sin^{3} x + \sin x = \cos x

\sin^{3} x + \sin x = \pm \sqrt{1 - \sin^{2} x}

\sin x = s

s^{3} + s = \pm \sqrt{1 + s^{2}}

squaring [might introduce false solutions]

and transforming

s^{6} + 2 s^{4} + 2 s^{2} - 1 = 0

{(s^{2})}^{3} + 2 {(s^{2})}^{2} + 2 (s^{2}) = 0

s^{2} = u - \frac{2}{3}

u^{3} + \frac{2}{3} u - \frac{47}{27} = 0

{Cardano}^{'} s solution

u = \sqrt[3]{\frac{47}{54} + \frac{\sqrt{249}}{18}} - \sqrt[3]{- \frac{47}{54} + \frac{\sqrt{249}}{18}}

it makes no sense to go on with this

u \approx 1.01987663

\Rightarrow

s^{2} = . 353209964

\Rightarrow

\sin x \approx \pm . 594314701

testing leads to

\sin x \approx . 594314701

\Rightarrow

x \approx 2 n π + 2.50517939 \lor x \approx 2 n π + . 636413265

Commented by mathdanisur last updated on 27/Sep/21

Very nice S er, thank you