Solve the following equations.
$|cosx|=cos(x+a).$

$|cosx|=cos(x+a)$

Now we use molecules

i.e.

We can write $cos(x+a)=|cosx|0<x<2\pi$

$cos(x+a)=cosx0<x<\pi /2$

$-cosx\pi /2<x<3\pi /2$

$cosx3\pi /2<x<2\pi$

when $0<x<\pi /2$ and $3\pi /2<x<2\pi$

Then $cos(x+a)=cosx$

Now using Trigonometry function

$co{s}^{-1}cos(x+a)=co{s}^{-1}cosx$

$x+a=x$

$a=0$

which is not possible

i.e. $\Rightarrow \pi /2<x<3\pi /2$

$\therefore cos(x+a)=-cosx\because$ we know that $-cosx=cos(-x)$

i.e. $cos(x+a)=cos(-x)$

Now we can write

$x+a=-x$

$2x=-a$

$x=-a/2$