# Math Help - Solving another simple trig. equation...

1. ## 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.

2. 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 $\theta$

3. another way:

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

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

then:

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

4. 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!

5. 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!

6. 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.

7. 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 $csc^2(\theta)=3\csc(\theta)$

8. 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 $\displaystyle \theta$...

9. 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!