Number of solutions of the equation: $\tan x=\cos2x$ in $[-π, π]$ I know that this question can be solved by using graphs, but in our examinations, we are not allowed to use calculators, or any digital devices.
I have a doubt that how to check whether the graphs will intersect or not at the points which are marked by red colour.
Can anyone suggest any simple method which is quick and easy which saves our time in the examination.
Any suggestion from your side will be appreciated.

 A: Here is an approach using polynomials. Write $\cos(2x)=\frac{1-\tan^2(x)}{1+\tan^2(x)}$. Then by multiplying both sides by $1+\tan^2(x)$, this is equivalent to solving the equation:
$$\tan(x)\big(1+\tan^2(x)\big)=1-\tan^2(x).$$
Let $u=\tan(x)$ to obtain the following polynomial equation:
$$u(1+u^2)=1-u^2\iff u^3+u^2+u-1=0.$$
The discriminant of this cubic polynomial is $-44$, so the equation has exactly one real root, which is a solution to the equation $\cos(2x)-\tan(x)=0$. By periodicity a solution occurs exactly once in every interval $[a,a+\pi]$, and in particular exactly twice in $[-\pi,\pi]$.
A: Hint: For $x\in[-\pi/2,-\pi/4)$ you have $\tan x<-1\le\cos 2x$. For $x\in(-\pi/4,0)$ you have $\tan x<0$ and $\cos 2x>0$.
A: Since the period of $\cos 2x$ is $\pi$, you only need to prove there are no solutions in the domain $(-\frac{\pi}{2}, 0)$.
Note that the minimum value of $\cos 2x$ is $-1$. So it is sufficient to to look for solutions where $-1 < \tan x \Rightarrow -\frac{\pi}{4} < x$. But at $x = -\frac{\pi}{4}$, $\cos 2x$ is $\cos \left( -\frac{\pi}{2} \right) = 0$. Since $\cos x$ and $\tan x$ is monotone increasing in the domain, $0 < \cos x < 1$, and $-1 < \tan x < 0$. Hence there are no roots in this domain.
There is a solution in the domain $[0, \frac{\pi}{2}]$ as the range of $\tan x$ is $[0, \infty]$, so it must intersect $\cos 2x$ which has a range of $[0, 1]$ by the Intermediate Value Theorem. Therefore, there is a solution once per period $\pi$. Since $[-\pi, \pi]$ contains two periods, there are hence two solutions in $[-\pi, \pi]$.
A: How about showing $\cos 2x > \tan x$ for $x\in \left[\frac{-\pi}{2},0\right]$ ?
That interval corresponds to the part where your first red mark appears. The inequality can be proved either by writing $\cos 2x$ in terms of $\tan x$ and showing that the cubic formed is always positive/negative (depending on the coefficient of $\tan ^3 x$) whenever $\tan x$ is negative.
Or
You can use a bit of calculus:
Define $f(x)=\cos 2x -\tan x$ and then see what happens to $f'(x)$ in $x\in \left[\frac{-\pi}{2},0\right]$ and I leave the rest up to you.
