Solve-for-real-numbers-4tan-x-sin-5x-cos-5-x-0-

Question Number 157964 by HongKing last updated on 30/Oct/21

Solve for real numbers :

4 \tan (x) + \frac{\sin (5 x)}{\cos^{5} (x)} = 0

Commented by tounghoungko last updated on 30/Oct/21

\sin (5 x) = 5 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x + \sin^{5} x

\Leftrightarrow \frac{4 \sin x \cos^{4} x + 5 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x + \sin^{5} x}{\cos^{5} x} = 0

\Leftrightarrow 9 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x + \sin^{5} x = 0

\sin x (9 \cos^{4} x - 10 \sin^{2} x \cos^{2} x + \sin^{4} x) = 0

(1) \sin x = 0 \Rightarrow x = {\begin{cases} 2 n π \\ (2 n + 1) π \end{cases}; n \in Z

(2) 9 {(\frac{1 + \cos 2 x}{2})}^{2} - \frac{5}{2} \sin^{2} 2 x + {(\frac{1 - \cos 2 x}{2})}^{2} = 0

l e t \cos 2 x = t

\Rightarrow 9 {(\frac{1 + t}{2})}^{2} - \frac{5}{2} (1 - t^{2}) + {(\frac{1 - t}{2})}^{2} = 0

\Rightarrow \frac{9 t^{2} + 18 t + 9}{4} - \frac{(5 - 5 t^{2})}{2} + \frac{1 - 2 t + t^{2}}{4} = 0

\Rightarrow 9 t^{2} + 18 t + 9 - 10 + 10 t^{2} + 1 - 2 t + t^{2} = 0

\Rightarrow 20 t^{2} + 16 t = 0

\Rightarrow 4 t (5 t + 4) = 0

(3) 4 t = 0 \Rightarrow 4 \cos 2 x = 0

\Rightarrow \cos 2 x = \cos \frac{π}{2}; x = \pm \frac{π}{4} + n π

(4) t = - \frac{4}{5} \Rightarrow \cos 2 x = - \frac{4}{5}

\Rightarrow 2 x = \pm \arccos (- \frac{4}{5}) + 2 n π

\Rightarrow x = \pm \frac{1}{2} \arccos (- \frac{4}{5}) + n π