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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.

Graph

UM Desai
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    I have removed the tags [tag:graph-theory] and [tag:random-graphs] because they are related to combinatorics. Please see the topic before adding a tag. – ShBh Aug 10 '20 at 02:33
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    If the calculator is not allowed then the problem has been designed that it can be done without a calculator. The graph of $cos2x$ is very easy to graph and $tanx$ is strictly increasing – imranfat Aug 10 '20 at 02:42
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    @imranfat That's not a justification. $x/10$ is also strictly increasing. – Andrei Aug 10 '20 at 02:46
  • @Andrei I too have same doubt – UM Desai Aug 10 '20 at 02:47
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    @Andrei Did you graph it by hand? I did. You don't need a graphing tool. We are not interested in the exact solutions, only in the number of solutions. It's really just a high school problem – imranfat Aug 10 '20 at 03:46

4 Answers4

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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 $[-\pi,\pi]$.

csch2
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    What is a discriminant of a cubic polynomial? can you explain to me? – UM Desai Aug 10 '20 at 02:52
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    It is a generalization of the discriminant for a quadratic polynomial. For a cubic polynomial $ax^3+bx^2+cx+d$, the discriminant is $b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$. If the discriminant is positive, there are three unique real solutions; if negative, one real solution; if zero, at least two equal solutions. – csch2 Aug 10 '20 at 02:54
  • Thanks for the information, it will be very helpful for me;) – UM Desai Aug 10 '20 at 02:56
  • you said that the solution occurs only once in [-π, π], but the graph shows two solutions. – UM Desai Aug 10 '20 at 03:02
  • @TobyMak both solutions are in this range, can you clarify what do you mean. you can see it in the graph also – UM Desai Aug 10 '20 at 03:05
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    The other solution is not in $[-\pi/2, \pi/2]$. You get one solution per period, but $\cos 2x, \tan x$ are periodic with respect to $\pi$. – Toby Mak Aug 10 '20 at 03:06
  • @TobyMak Ok got it – UM Desai Aug 10 '20 at 03:13
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    By periodicity, a solution occurs ever $[a,a+\pi]$ because the function $\cos(2x)-\tan(x)$ has period $\pi$ not $2\pi$. – Mars Aug 10 '20 at 03:50
  • @Morph Thanks for the correction, I've edited the post – csch2 Aug 10 '20 at 14:26
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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 $\cos 2x>0$.

Andrei
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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 $[-\pi, \pi]$.

Toby Mak
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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.

AryanSonwatikar
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