Question

Find the absolute maximum and minimum of function $y=2\cos 2x-\cos 4x$.

Answer 1

Given $y=2\cos 2x-\cos 4x$

$\Rightarrow \frac{dy}{dx}=-4\sin 2x+4\sin 4x$

For maxima or minima

$\frac{dy}{dx}=0$

$\Rightarrow 4(\sin 4x-\sin 2x)=0$

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

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

$\Rightarrow \sin 2x=0$ or $2\cos 2x-1=0$

Now, $\sin 2x=0\Rightarrow 2x=0,\pi ,2\pi \Rightarrow x=0,\frac{\pi }{2},\pi$

$2\cos 2x-1=0\Rightarrow \cos 2x=\frac{1}{2}\Rightarrow 2x=\frac{\pi }{3},\frac{5\pi }{3}\Rightarrow x=\frac{\pi }{6},\frac{5\pi }{6}$

Min value $=-3$

Max value $=\frac{3}{2}$