# Math Help - Solving a relatively easy trig. equation...

1. ## Solving a relatively easy trig. equation...

Hi!

The equation is...

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

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

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

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

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?

2. edit: my algebra sucked

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

0 doesn't have a sign so squaring won't make a difference and you can simply solve $\sin \theta = 0$

3. 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?

4. From the graph of sin(x)

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

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

5. ok, i get it.

thank you.

6. 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.

7. Think of the unit circle. As you move along the unit circle, at which angles does the vertical distance become 0?

8. 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.**