Finding maximum and minimum values of $\sin(2x) - x$ for $x\in\left[-\frac \pi 2, \frac \pi 2\right]$

Question

This is a problem that I wasn't able to solve. Please help.

Find the maximum and minimum values of the function: $$y = \sin(2x) - x$$ where domain of $x$ is $\left[-\frac \pi 2, \frac \pi 2\right]$.

Answer 1

First differentiate $y=\sin(2x)-x$.
You get $y^\prime=2\cos (2x)-1.$ At an extrema this value will be zero. So equate the above equation to zero and solve for $x$. Remember the value of $x$ has to lie in the given interval. It can be seen easily that the $x=\frac{\pi}{6}$.

Then find the second derivative of $y$.

Substitute this value of $x$ obtained.

If the resulting value is positive then the function has a minima at that $x$.
If it is negative then the function has a maxima.