Solve sin^2(θ) = 3cos^2(θ)
What I have gotten so far is:
sin^2/cos^2=3
tan^2=3
x= arctan(sqrt(3))
x=pi/3
I know tan(2pi)=0 so I have the solution of pi/3+2piN. This is reflected in multiple of the choices. How do I get the rest of the equations?
Also, one of the possible answers is "not one of these choices"

Question

Solve sin^2(θ) = 3cos^2(θ)
What I have gotten so far is:
sin^2/cos^2=3
tan^2=3
x= arctan(sqrt(3))
x=pi/3
I know tan(2pi)=0 so I have the solution of pi/3+2piN. This is reflected in multiple of the choices. How do I get the rest of the equations?
Also, one of the possible answers is "not one of these choices"

myininaya · Answer

ISe the identity that sin^2(theta)=1-cos^2(theta)
do not do any divisions as you have done above because cos can be 0 sometimes

myininaya · Answer

USE *

triciaal · Answer

solve    aren't  you just suppose to find the angle?
tan^2 theta = 3 so tan theta = sqrt 3
is x the radius of the circle?
are you finding the equation of the circle with that angle?

anonymous · Answer

The possible answers are :
a: pi/3, 5pi/3
b: pi/3+2pin, 2pi/3+2pin, 4pi/3+2pin, 5pi/3+2pin
c:  pi/3+2pin,5pi/3 + 2pin
d: none of these

triciaal · Answer

cos theta = 1/2, -1/2

myininaya · Answer

@sccitesla As I was saying you do not won't to divide the initial equation by cos
because cos can sometimes be 0 (and you do not divide by 0)
If you do, this could eliminate some answers that you should have had

myininaya · Answer

Use the identity mentioned above 1-cos^2(theta)=sin^2(theta)

anonymous · Answer

How can i use that density with no stray additions. Am i supposed to add them in?

myininaya · Answer

I see sin^2(theta) so I would replace sin^2(theta) with 1-cos^2(theta)
like so:
$$1-\cos^2(\theta)=3\cos^2(\theta) $$

myininaya · Answer

can you solve this?

anonymous · Answer

...sorry brain fart.

anonymous · Answer

im still in the same place i was before i believe...... though hopefully i missed something

anonymous · Answer

arccos(.5)=pi/3

anonymous · Answer

+-.5=cos(x)
cos(2pi/3)= -.5
cos(2pi/3)= -.5
cos(4pi/3)= -.5
cos(5pi/3)= .5

anonymous · Answer

no answers have been eliminated

myininaya · Answer

I'm going to add cos^2(theta) on both sides giving me
$$1=4 \cos^2(\theta)$$

divide both sides by 4
$$\frac{1}{4}=\cos^2(\theta)$$
Now to get rid of the square you must square root both sides

this will give you two equations to solve.

myininaya · Answer

Oh I think you got it right there
1/2=cos(x) or -1/2=cos(x)

myininaya · Answer

@sccitesla You found your answer.

myininaya · Answer

You have already stated it.

anonymous · Answer

Those are not in the choices

anonymous · Answer

nvm

myininaya · Answer

so you don't see pi/3, 2pi/3, 4pi/3, 5pi/3
all of those +2npi

anonymous · Answer

i see. Stupid freaking things like this is what gets me stuck -_-

anonymous · Answer

wait... one second

anonymous · Answer

so the only answer that has solutions in both positive and negative is b. so that must be the answer. Ok thanks