Trigonometric inequality $ 3\cos ^2x \sin x -\sin^2x <{1\over 2}$

The solution and the problem do not match. If you define:

$$f(x)=3\cos ^2x \sin x -\sin^2x$$

You would expect to see:

$$f(\pi/18)=1/2$$

Actually it's 0.475082. So the problem and the solution do not match. But if you edit the problem just a little bit:

$$ 3\cos ^2x \sin x -\sin^3x <{1\over 2}$$

...the solution and the problem seem to be matching (note the cube instead of square in the second term on the left).

So we have a typo here! :) And the correct version of the problem is likely easier.