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?
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?
$2+\cot\theta=\csc\theta\implies 2+\frac{\cos\theta}{\sin\theta}=\frac{1}{\sin\theta}$
$\implies2\sin\theta+\cos\theta=1$ $(\sin\theta\neq 1)$
$\implies \frac{2}{\sqrt 5}\sin\theta+\frac{1}{\sqrt 5}\cos\theta=\frac{1}{\sqrt 5}$
$\implies \cos\alpha\sin\theta+\sin\alpha\cos\theta=\frac{1}{\sqrt 5}$ 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 $\theta= \pi-\arctan(4/3)$
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.
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 $(\frac {\pi}2,\pi)$
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.