Let ${\displaystyle f:R\to R}$ be defined by ${\displaystyle f\left(x\right)=\cos (5x+2)}$. Is f invertible? Justify your answer.

For invertible of f, f must be bijective (i.e, one-one onto).
If ${\displaystyle x_1,x_2\in R}$,
then ${\displaystyle f\left(x_1\right)=f\left(x_2\right)}$
${\displaystyle \Rightarrow \cos (5x_1+2)=\cos (5x_2+2)}$
${\displaystyle \Rightarrow 5x_1+2=2n\pi \pm (5x_2+2)}$
${\displaystyle \Rightarrow x_1\ne x_2}$
${\displaystyle \therefore }$ f is not one-one.
But ${\displaystyle -1\le \cos (5x+2)\le 1}$
${\displaystyle \therefore -1\le f\left(x\right)\le 1}$
Range ${\displaystyle =[-1,1]\subset R}$
${\displaystyle \therefore }$ f is into mapping.
Hence, the function f(x) is no bijective and so it is not invertible.