Trigonometric equation solutions. For $0<\theta<\pi/6$ all the values of the expression
$\tan^23\theta \cos^2\theta-4\tan3\theta \sin2\theta+16\sin^2\theta$ lies in what interval.
I actually took $\sin^2\theta$ common out of the expression to yield,
$$(\frac{\tan^23\theta}{\tan^2\theta}-\frac{8\tan3\theta}{\tan\theta} +16)\sin^2\theta$$
$$=(\frac{\tan3\theta}{\tan\theta} -4)^2\sin^2\theta$$
Now how can I now proceed further ahead.

$$(\frac{tan3\theta}{tan\theta} -4)^2sin^2\theta$$
$$=(\frac{11tan^2\theta-1}{1-3tan^2\theta} )^2sin^2\theta$$
Now $sin$ is an increasing function in the domain. 
Also if you check the derivative(s) (or rough graphs of$\frac{11tan^2\theta-1}{1-3tan^2\theta}$) you will notice that $\frac{11tan^2\theta-1}{1-3tan^2\theta}$ is symmetric about $0$ (and value at $\theta=0$ is $-1$ ) and has a root at (say) $0\lt x_0\lt \pi/6$  (at $\theta=\pi/6 $ its $\infty$). So squaring the thing will just reflect the part below $X$ axis above , keeping the root intact . So minimum value of the expression is zero (achieved at 2 points $0\  \&\  x_0$) and maximum is infinity 
