# Solving another simple trig. equation...

• Feb 17th 2011, 04:44 PM
jonnygill
Solving another simple trig. equation...
The equation is...

$1+cot^2\Theta-3csc\Theta=0$

I tried substituting $\dfrac{cos^2\Theta}{sin^2\Theta}$ for $cot^2\Theta$ and replaced $csc\Theta$ with $\dfrac{1}{sin\Theta}$.

I was left with...

$\dfrac{cos^2\Theta}{sin^2\Theta}-\dfrac{3}{sin\Theta}=-1$

And now i'm pretty confused. I tried cross multiplying and getting a common denominator, but that just kept me going in circles. Which is when i realized i was doing something wrong.
• Feb 17th 2011, 04:57 PM
e^(i*pi)
Good thus far. Now multiply through by $\sin^2 \theta$ (which happens to be the LCD of sin^2(x), sin(x) and 1)

$\cos^2\theta- 3\sin \theta +1 = 0$

Now you can sub in $\cos^2 \theta = 1- \sin^2 \theta$

$1-\sin^2 \theta - 3\sin \theta +1 = 0 \implies \sin^2 \theta + 3\sin \theta -2 = 0$

(I moved everything to the other side because I prefer dealing with positive values of a)

Use the quadratic formula to solve for $\sin \theta$ and then $\theta$
• Feb 17th 2011, 05:18 PM
harish21
another way:

use the identity: $\csc^2(x)-\cot^2(x)=1$

$\therefore 1+\cot^2(x)=\csc^2(x)$

then:

$1+\cot^2(\theta)-3\csc(\theta)=0 \implies \csc^2(\theta)=3\csc(\theta)$
• Feb 17th 2011, 05:23 PM
jonnygill
oh, cool.

ok, so i plugged into the quad. equation the numbers 1, 3, and -2 respectively.

when i used the positive square root of 17 i got an answer of about 34.1633 degrees.

However, when i used the negative square root of 17 i got an error on my calculator. does this simply mean there is only one answer?

thanks!
• Feb 17th 2011, 05:28 PM
e^(i*pi)
You won't get an answer because it's an extraneous solution since the range is $-1 \leq \cos \theta \leq 1$. You can discard that answer that gave you an error.

I suspect there is more than one solution of 0 < x < 360

what they are I don't know since it's half 2 in the morning and I should be in bed!
• Feb 17th 2011, 05:38 PM
jonnygill
thank you.

if there is another solution with 0 < x < 360 i do not know what it is.

if anyone knows that there is another solution, how to find it, and is willing to show me i would be very grateful.
• Feb 17th 2011, 05:51 PM
jonnygill
Quote:

Originally Posted by harish21
another way:

use the identity: $\csc^2(x)-\cot^2(x)=1$

$\therefore 1+\cot^2(x)=\csc^2(x)$

then:

$1+\cot^2(\theta)-3\csc(\theta)=0 \implies \csc^2(\theta)=3\csc(\theta)$

thanks!

but i can't use the quad. equation for the csc function right? so how would i solve $csc^2(\theta)=3\csc(\theta)$
• Feb 17th 2011, 06:26 PM
Prove It
Sure you can...

$\displaystyle \csc^2{\theta} - 3\csc{\theta} = 0$

$\displaystyle \csc{\theta}(\csc{\theta} - 3) = 0$

$\displaystyle \csc{\theta} = 0$ or $\displaystyle \csc{\theta} - 3 = 0$.

Solve each for $\displaystyle \theta$...
• Feb 17th 2011, 07:32 PM
jonnygill
Quote:

Originally Posted by Prove It
Sure you can...

$\displaystyle \csc^2{\theta} - 3\csc{\theta} = 0$

$\displaystyle \csc{\theta}(\csc{\theta} - 3) = 0$

$\displaystyle \csc{\theta} = 0$ or $\displaystyle \csc{\theta} - 3 = 0$.

Solve each for $\displaystyle \theta$...

cool!

so i don't even need the quadratic equation for this problem.

thanks!