# How would I find the solution to $\sin x= \frac{1}{\sqrt{5}}$ manually?

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

$$\frac{\cos(x)}{\sin(x)}=2$$

$$\cos(x)=2\sin(x)$$

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

$$\cos^2(x)=4\sin^2(x)$$

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 $$\sin(x)$$.....

• 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? Commented Jan 24, 2023 at 14:01
• 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 $[0, \pi]$. Commented Jan 24, 2023 at 14:02
• 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? Commented Jan 24, 2023 at 14:04
• From $\sin^2 x=\frac15$, you can't deduce $\sin x=1/\sqrt 5$. Do you see why? Commented Jan 24, 2023 at 14:31
• @Vasili: I think you must mean $(0,\pi)$. Commented Jan 24, 2023 at 14:36

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 $$$$y = \sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$$$

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

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

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.