How to solve 2 - 2sin^2(theta) = cos(theta) on the domain [0,2pi]

[aisha]

2-2\sin^2(\theta)=\cos (\theta)

I need to solve this quadratic trig equation algebraically of the domain 0<=x<=2pi I know how to do that but in order to solve the trig functions have to be the same I'm having a little trouble how do I change the cos into sin? It's not cos^2 so we can't use the pythagorean identity.

 

[dextercioby]

Factor that 2 and use the fundamental identity
\sin^{2}x+\cos^{2}x =1

Daniel.

 

[aisha]

Oh thanks here's what I got

2(1-\sin^2\theta)=\cos \theta
2(\cos^2\theta)=\cos\theta

Expand 2\cos^2\theta=\cos\theta

Rearrange 2\cos^2-\cos\theta=0

 

[dextercioby]

Factor cos\theta and end up with 2 simple equations...

Daniel.

 

[aisha]

\cos \theta=0 and \cos\theta=1/2

 

[dextercioby]

Perfect.Sove each of them and then write the final solution to the initial eq.

Daniel.

 

[aisha]

I got 300 degrees and 60 degrees as solutions for cos (x)= 1/2

but I am not sure for cos(x)=0

its either 0 degrees and 180 degrees, or 90 degrees

 

[BobG]

Try 90 and 270.

Cos (0) is 1
Cos (180) is -1

 

[aisha]

ok so are there 4 solutions

90 degrees 270 degrees, 300 degrees and 60 degrees

Is that correct?

 

[dextercioby]

Yes,u can make a graph (trigon.circle) to check the validity.


Daniel.

 

[aisha]

but my answers are right?

 

[dextercioby]

Yes,they are...And what's so funny...?My new haircut...?

Daniel.

 

[aisha]

lol thanks sooooooooooooo much i finally got something that's the funny part!