trig help with equation 2cot2x + cot^2x = 1

[java908]

i need help with this equation: 2cot2x+cot^2x=1

i know that cot=cos/sin so i think its supposed to be like this 2(cos2x/sin2x)+cos2x/sin2x=1
Im not sure about the last one since its cot^2x

i get stuck and dont know how to proceed, i dont know a lot about cot and how to use it, i tried to read some articles but it dosent help me much, i would be really thankful if someone could guide me through this or try to explain for me. thanks.

 

[Loren]

Re: trig help!

Using the information you have stated I would get...
$\displaystyle 2(\frac{\cos 2x}{\sin 2x})+\frac{\cos^2x}{\sin^2x}=1$

If I were doing it this way, I would try to express the cos2x and sin2x in terms of cos x and sin x and go from there.

However, I think you will be better off by converting cotangents to equivalent "tangents" and solve the equation in terms of tangents. In other words, don't involve sine and cosine. I got as far as (tan x + 1)(tan x + 1)(tan x - 1) = 0. From there, it is duck soup.

 

[soroban]

Re: trig help!

Hello, java908

$\displaystyle \text{Solve: }\;2\cot2x + \cot^2\!x\:=\:1$

I'd write everything in terms of $\displaystyle 2x$

$\displaystyle \text{The left side is: }\:2\,\frac{\cos2x}{\sin2x} + \frac{\cos^2\!x}{\sin^2\!x} \;=\;2\,\frac{\cos2x}{\sin2x} + \frac{\frac{1+\cos2x}{2}}{\frac{1-\cos2x}{2}} \;=\;2\,\frac{\cos2x}{\sin2x} + \frac{1+\cos2x}{1-\cos2x}$

$\displaystyle \text{And we have: }\:2\,\frac{\cos2x}{\sin2x} + \frac{1 + \cos2x}{1-\cos2x} \;=\;1$


$\displaystyle \text{Multiply by }\sin2x(1 - \cos2x):$

. . $\displaystyle 2\cos2x(1-\cos2x) + \sin2x(1+\cos2x) \;=\;\sin2x(1-\cos2x)$

. . $\displaystyle 2\cos2x - 2\cos^2\!2x + \sin2x + \sin2x\cos2x \;=\;\sin2x - \sin2x\cos2x$

. . $\displaystyle 2\cos2x - 2\cos^2\!2x + 2\sin2x\cos2x \;=\;0$


$\displaystyle \text{Factor: }\;2\cos2x(1 - \cos2x + \sin2x) \;=\;0$


$\displaystyle \text{Then solve the two equations:}$

. . $\displaystyle 2\cos2x \:=\:0$

. . $\displaystyle 1 - \cos2x + \sin 2x \:=\:0$