What is $k_{\text{max}}$? 
If
$[1-\cos x][1 - \cos 2x][1 - \cos 3x] = k\ ; 0º < x < 90º$
Find $k_{\text{max}}$

I have no idea how to solve this 
I've got  $8\left[\sin\left(\frac{x}{2}\right)\times\sin x \times \sin\left(\frac{3x}{2}\right)\right]^2 = k$
Help me please ...
Thank you to all.

$$\Big(2\sin\dfrac{x}{2}\sin x\sin\dfrac{3x}{2}\Big)^2=\sin^2 x\Big(\cos x-\cos 2x\Big)^2=(1-\cos^2x)\Big(-2\cos^2x+\cos x+1\Big)^2$$ Let $$f(t)=(1-t^2)(2t^2-t-1)^2=(1-t)^3(1+t)(1+2t)^2$$ where $t\in[0,1]$ Try to find the maximum of $f.$ I think that would be enough.
