2cos-2-x-25-cot-x-0-x-

Question Number 119220 by benjo_mathlover last updated on 23/Oct/20

2 \cos^{2} x - 25 \cot x = 0

x = ?

Answered by MJS_new last updated on 23/Oct/20

\cos x = \frac{1}{\sqrt{1 + \tan^{2} x}}

let t = \tan x

\frac{2}{t^{2} + 1} - \frac{25}{t} = 0

t^{2} - \frac{2}{25} t + 1 = 0

t = \frac{1}{25} \pm \frac{4 \sqrt{39}}{25} i

no real solution

x = n π + \arctan \frac{1 \pm 4 \sqrt{39} i}{25}

Commented by MJS_new last updated on 23/Oct/20

I forgot \lim_{t \to \pm \infty} (\frac{2}{t^{2} + 1} - \frac{25}{t}) = 0

\Rightarrow x = n π \pm \frac{π}{2}

Commented by benjo_mathlover last updated on 23/Oct/20

y e s

Answered by 1549442205PVT last updated on 23/Oct/20

2 \cos^{2} x - 25 \cot x = 0

\Leftrightarrow {2 \cos}^{2} x - 25 \frac{cosx}{sinx} = 0 \Rightarrow sinx \neq 0 \Leftrightarrow x \neq k π

\Leftrightarrow cosx (2 sinxcosx - 25) = 0

\Leftrightarrow cosx (\sin 2 x - 25) = 0 (since \sin 2 x - 25) \neq 0

\Leftrightarrow cosx = 0 \Leftrightarrow x = \frac{π}{2} + m π (m \in Z)

Commented by MJS_new last updated on 23/Oct/20

you are right . but {it}^{'} s even shorter

\cos x (2 \cos x - \frac{25}{\sin x}) = 0 \Rightarrow \cos x = 0

Commented by 1549442205PVT last updated on 23/Oct/20

But i want to explain that \sin 2 x - 25 = 0

has no real roots, Sir