Substitute x in the place of

Multiply each side by negative one

Now use quadratic formula

\ \

\ \

But the range of sinx is [ 1 , -1 ] we can have only

\ \ \ \

\ \ \ \ By the second derivative it is locally minimum at

Substitute in given f(x)

\ \

Now check the end points f(0) = -5 and

\ \ = -2.595

So the maximum value of  

= -1.517

And the minimum value is f(o) = -5