I'm laboring through my trigonometric course and I get stuck at the following problem:

"How many solution do we have for this equation in the range $[-5\pi, 3\pi]$ $$cot(x)=2$$

I figure this must mean that



I don't really know any trigonometric identity that applies, but I try taking the square of both sides:


If we use the fact that $\sin^2(x)+\cos^2(x)=1$ then:

$$1-\sin^2(x)=4\sin^2x$$ $$1=5\sin^2(x)$$ $$\frac{1}{5}=\sin^2(x)$$ $$\frac{1}{\sqrt{5}}=\sin(x)$$

This really doesn't help much though. I'm supposed to figure out the solution for $x$, so the angle. I'm supposed to answer every question without using a calculator and I can't really see how I can arrive at $\frac{1}{\sqrt{5}}$ using any of the known identities for $\sin(x)$.....

  • 1
    $\begingroup$ You don't have to solve this equation manually, or numerically, in order to figure out how many solutions there are. Try drawing a sketch of $\sin^2 x$ in the given range. How many times does it take each value? $\endgroup$ Jan 24 at 14:01
  • 1
    $\begingroup$ You only need to find the number of solutions. Do you know how to sketch the graph $y= \cot x$? Sketch the graph then draw the line $y=2$ and count the number of intersections. You can also use that cotangent is periodic and bijective on $[0, \pi]$. $\endgroup$
    – Vasili
    Jan 24 at 14:02
  • 2
    $\begingroup$ You're probably more familiar with the graph of $y=\tan x$. The given equation is equivalent to $\tan x= \tfrac12$. How many times does the line $y=\tfrac12$ intersect the graph of $y=\tan x$ in that range? $\endgroup$ Jan 24 at 14:04
  • $\begingroup$ From $\sin^2 x=\frac15$, you can't deduce $\sin x=1/\sqrt 5$. Do you see why? $\endgroup$
    – TonyK
    Jan 24 at 14:31
  • $\begingroup$ @Vasili: I think you must mean $(0,\pi)$. $\endgroup$
    – TonyK
    Jan 24 at 14:36

2 Answers 2


This goes off topic, but it is possible to find a solution for $\sin(x)=1/\sqrt{5}$ for (e.g.) $x$ close to 0 by series inversion by hand.

Starting with the expansion \begin{equation} y = \sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+... \end{equation}

and assuming that $x = a_0y + a_1y^3 + a_2y^5$, inserting this into the above equation gives that

\begin{equation} y = a_0y+y^3\left(a_1 - \frac{a_0^3}{3!}\right) + y^5\left(a_2 - \frac{3a_0^2a_1}{3!} + \frac{a_0^5}{5!} \right) + O(y^7) \end{equation}

Solving for the coefficients gives $x = y+\frac{y^3}{6}+\frac{3y^5}{40} + O(y^7)$. Now, setting $y=\sin(x)=1/\sqrt{5} \approx 1/2.25$ gives that one solution to $\sin(x)=1/\sqrt{5}$ is approximately $0.4603...$. The exact is $0.4636...$ which is quite close.


We want to find $x$ in degrees for which $$ \sin{x} = \frac{1}{\sqrt{5}} \hspace{10mm} (1) $$

So easiest way to find $x$ is we know that from $\sin{x}$ is ratio of perpendicular over hypotenuses in . So from $(1)$ we know that a right angle triangle whose perpendicular is 1, hypotenuses is $\sqrt{5}$ and base is 2. So with the help of ruler make a line in $x-axis$ of 2 units and in $y-axis$ 1 units. Then make a line joining both of these points $(2,0) \And (0,1)$. And then with the help of protractor measure the angle between $x-axis$ and that line we just made joining the 2 points.


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