Question

Find the value of $x$ for which the derivative of the function $f\left(x\right)=20\cos 3x+12\cos 5x-15\cos 4x$ is equal to zero?

Answer 1

$f\left(x\right)=20\cos 3x+12\cos 5x-15\cos 4x$

$f^{\prime }\left(x\right)=20(-\sin 3x)\left(3\right)+12(-\sin 5x)\left(5\right)+15\sin 4x\left(4\right)$

$=-60\sin 3x-60\sin 5x+60\sin 4x$

$=60[\sin 3x-\sin 3x-\sin 5x]=0$

$\Rightarrow \sin 4x=\sin 3x+\sin 5x$

$\Rightarrow \sin 4x=2\sin \frac{8x}{2}\cos \frac{2x}{2}$

$\sin 4x=2\sin 4x\cos x$

$\sin 4x(1-2\cos x)=0$

$\therefore \sin 4x=0$ and $1-2\cos x=0$

$x=\frac{\pi m}{4}$

and $\cos x=\frac{1}{2}$

$x=\pm \frac{\pi }{3},\frac{-5\pi }{3},\frac{7\pi }{4}...$

$x=\pi \left[\frac{6n+1}{3}\right]$

where $m,nϵZ$ (integers).

$\therefore x=\frac{\pi m}{4},\frac{\pi (6n+1)}{3},|m,nϵZ$