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.