Trigonometry! #29

Both the LHS and RHS are positive, therefore AM-GM inequality applies.

We note that the $RHS = x^2 + \dfrac {1}{x^2} \ge 2$

Now, consider the maximum value of LHS:

$\quad 2\cos^2{\frac{x}{2}}\sin^2{x} = (\cos{x}+1)(1 - \cos^2{x}) = 1 + \cos{x} - \cos^2{x} - \cos^3{x}$

$\Rightarrow \dfrac {d}{dx} (2\cos^2{\frac{x}{2}}\sin^2{x}) = -\sin{x} + 2\cos{x}\sin {x} + 3\cos^2{x}\sin{x}$

$\quad \dfrac {d}{dx} (2\cos^2{\frac{x}{2}}\sin^2{x}) = 0\quad \Rightarrow \sin{x} = 0$ or $3\cos^2{x} + 2\cos{x} - 1 = 0$

When $\sin{x} = 0\quad \Rightarrow LHS = 0$ hence not the maximum.

When $3\cos^2{x} + 2\cos{x} - 1 = 0\quad \Rightarrow (3\cos {x} -1)(\cos{x} +1) = 0$ since $\cos{x} > 0\quad \Rightarrow \cos{x} = \frac {1}{3}$

Therefore, the maximum value of:

$\quad 2\cos^2{\frac{x}{2}}\sin^2{x} = 1 + \cos{x} - \cos^2{x} - \cos^3{x} = 1 + \frac {1}{3} - (\frac {1}{3})^2 - (\frac {1}{3})^3$

$\quad \quad \quad \quad \quad \quad \quad \quad = 1 + \frac {1}{3} - \frac {1}{9} - \frac {1}{27} = \frac {32}{27} < 2$

It can be seen that LHS is always less than RHS, therefore there is $\boxed{0}$ solution.