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.

• Personally, I think you should just find sup and inf of that function on $\left(0,\frac\pi6\right)$ and see if they are is attained for some $\theta$. – user228113 Jun 13 '16 at 18:29

$$(\frac{tan3\theta}{tan\theta} -4)^2sin^2\theta$$ $$=(\frac{11tan^2\theta-1}{1-3tan^2\theta} )^2sin^2\theta$$
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