# Trigonometric equations: $2 + \cot\theta = \csc\theta$

Solve

$$2+\cot\theta = \csc\theta$$ where $$0 \leq \theta \lt 2\pi$$

The suggested answer is $2.21$ only (in rad, corr to $3$ sig. fig.)

My reasonable guess is there are at least two solutions. Any suggestion?

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I suggest that you start by graphing the function $y=2+\cot x - \csc x$. Or wolfram alpha is your friend. wolframalpha.com/input/?i=plot+2%2B+%28cot+x%29-%28csc+x%29 –  sabertooth Sep 28 '12 at 11:03

Notice $$2+\cot\theta=\csc\theta\Leftrightarrow 2+\frac{\cos\theta}{\sin\theta}=\csc\theta\Leftrightarrow \csc\theta(2\sin\theta+\cos\theta)=\csc\theta$$

$\csc\theta$ is never zero, so look at $2\sin\theta=1-\cos\theta$

How many solutions does this have in $(0,2\pi)?$ Perhaps picturing the behaviour of the RHS and LHS alone will help convince you that there is only one solution. Or you simply solve it as Avatar has suggested.

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thanks, but why 0.927 can't be a solution? –  Paul Smith Sep 28 '12 at 12:27
@PaulSmith could you explain where you are getting that value from? –  user39572 Sep 28 '12 at 12:38
I applied pythargeon identitites $$1+cot^2\theta=csc^2\theta$$ as suggested by lab bhattacharjee below. I got $$sin\theta=\frac{4}{5}$$ –  Paul Smith Sep 28 '12 at 12:40
@PaulSmith The problem is you are implicitly generating another solution by considering the identity: but the additional solution doesn't comply with the original equation. Sort of like saying $x=1\Rightarrow x^2=1\Rightarrow x=1$ or $-1$, when $x\neq -1$. $\theta=\arcsin(\frac{4}{5})$ is not a solution. –  user39572 Sep 28 '12 at 13:04

We know $\csc^2\theta-\cot^2\theta=1$

$\implies (\csc\theta+\cot\theta)(\csc\theta-\cot\theta)=1$

Given $\csc\theta-\cot\theta=2$

So, $\csc\theta+\cot\theta=\frac 12$

So, $2\csc\theta=\frac 5 2\implies \sin\theta=\frac 4 5>0$

$\cot\theta=\frac 12- \csc\theta=\frac 12- \frac 5 4<0$

So, $\theta$ will lie in the 2nd quadrant, the principal value in $(\frac {\pi}2,\pi)$

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$2+\cot\theta=\csc\theta\implies 2+\frac{\cos\theta}{\sin\theta}=\frac{1}{\sin\theta}$

$\implies2\sin\theta+\cos\theta=1$ $(\sin\theta\neq 1)$

$\implies \frac{2}{\sqrt 5}\sin\theta+\frac{1}{\sqrt 5}\cos\theta=\frac{1}{\sqrt 5}$

$\implies \cos\alpha\sin\theta+\sin\alpha\cos\theta=\frac{1}{\sqrt 5}$ where $\alpha=\arctan(\frac{1}{2})$

$\implies \sin(\alpha+\theta)=\frac{1}{\sqrt 5}$

I think you can solve it from here.

Alternatively,

Let $\cot\theta=x,$ then, $\csc^2\theta=1+\cot^2\theta=1+x^2$

Thus, $(2+\cot\theta)^2=\csc^2\theta$

$\implies 4+x^2+4x=x^2+1$

$\implies 4x=-3\implies x=-3/4\implies \tan\theta=-4/3\implies \theta=\pi-\arctan(4/3),2\pi-\arctan(4/3)$

But only $\pi-\arctan(4/3)$ satisfies the original equation.Thus, the solution is $\theta= \pi-\arctan(4/3)$

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ya, we can do it with either compound angles formula or algebraic substitution. I just wonder why we have to reject $$\theta = sin^{-1}\frac{4}{5}$$ in the first quadrant, please kindly hint. –  Paul Smith Sep 28 '12 at 12:36
yes, we are getting $\sin\theta=4/5$, but putting that in equation we are also getting $\cos\theta=-3/5$ $\sin\theta +, \cos\theta -$ happens only in second quadrant. –  Aang Sep 28 '12 at 12:41
if you substitute $\theta=\sin^{-1}(4/5)$ in first quadrant , then, $2\sin\theta+\cos\theta=2(4/5)+(3/5)=11/5\neq 1$ as we wanted –  Aang Sep 28 '12 at 12:43
oh thanks! I got that now –  Paul Smith Sep 28 '12 at 12:44
may i ask how do you realize that at first glance? –  Paul Smith Sep 28 '12 at 12:44

Square both sides and we have

$$4+4\cot\theta+\cot^2\theta=\csc^2\theta$$

Using the Pythagorean identity for $\csc$ and $\cot$ gives $$4\cot\theta=-3$$ From here, it's a simple matter to find two possible solutions in $(-\pi,\pi]$: $\arctan(-4/3)$ and $\arctan(-4/3)+\pi$. Since we squared both sides at one point, we may have introduced extraneous solutions. The first of these angles, $\arctan(-4/3)$ is in quadrant IV, so its $\csc$ is negative. Meanwhile, $2+\cot(\arctan(-4/3))=2-3/4$ is positive. So the original equation cannot be satisfied by $\arctan(-4/3)$. For a final confirmation, check that $\arctan(-4/3)+\pi$ does indeed satisfy the original equation.

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