# Solving a relatively easy trig. equation...

• Feb 17th 2011, 04:05 PM
jonnygill
Solving a relatively easy trig. equation...
Hi!

The equation is...

$\displaystyle 2cos^2\Theta=2+sin^2\Theta$

i'm trying to solve for theta, limiting domain from 0 to 360 degrees.

After substituting $\displaystyle 1-sin^2\Theta$ for $\displaystyle cos^2\Theta$ and simplifying i ended up with...

$\displaystyle -3sin^2\Theta=0$ OR $\displaystyle sin\Theta=\sqrt0$

after applying arcsine i came up with an answer of 0 degrees. Now, my graphing calculator of course does not provide ALL possible answers. I noticed that 180 degrees is also an answer. I guess this is because of the theoretical -0 degrees that when squared would result in 0 too?

Did i do this correctly?
• Feb 17th 2011, 04:07 PM
e^(i*pi)
edit: my algebra sucked

$\displaystyle 2 - 2\sin^2 \theta = 2 + \sin^2 \theta \rightarrow 3\sin^2 \theta = 0$

0 doesn't have a sign so squaring won't make a difference and you can simply solve $\displaystyle \sin \theta = 0$
• Feb 17th 2011, 04:25 PM
jonnygill
ok this makes sense. but, 180 degrees is also an answer. since sin(180)=0. So, there are two answers. One the calculator gives me. How do i arrive at the fact that 180 degrees is also an answer?
• Feb 17th 2011, 04:31 PM
e^(i*pi)
From the graph of sin(x)

$\displaystyle \sin x = 0 \right \text{ hence } x = 0 + 180n \text{ where } n \in \mathbb{Z}$

Hence 360 is a potential solution depending on your domain.

See the wolfram graph of sin(x): http://www.wolframalpha.com/input/?i...+from+0+to+2pi
• Feb 17th 2011, 04:50 PM
jonnygill
ok, i get it.

thank you.
• Feb 17th 2011, 07:44 PM
jonnygill
actually that wasn't entirely true. I understand that since the graph intersects the x-axis at 0 pi and pi that these are two possible solutions.

what i don't understand is how i would arrive at 180 degrees or pi as also being a solution.
• Feb 17th 2011, 07:49 PM
Prove It
Think of the unit circle. As you move along the unit circle, at which angles does the vertical distance become 0?
• Feb 17th 2011, 08:29 PM
jonnygill
ok. so sin function gives you the y-coordinate and cos function gives you the x-coordinate. and tan gives you a ratio of the two.

all within the unit circle of course.

that is, when you plug in degrees.**