fairly easy Trig Problem?

Question

Im pretty weak at math and need to know how to solve this I already know the answer I just need to know how they got there

1/3=cos^2 theta

theta comes out to be 35.3, but I need to know how they got there, thanks

the answer i placed up was actually for 1/3=1/2cos^2(theta)

Answer 1

Ok, I can give you two methods, the first one :

1/3 = cos^2(theta)

applying sqrt :

0.57 = cos(theta)

theta = arccos(0.57)

theta = 54.7

Another way to do it :

1/3 = cos^2(theta)

2/3 = 2*cos^2(theta)

remembering : 2cos^2(theta) - 1 = cos(2theta)

1 + cos(2theta) = 2/3

cos(2theta) = -1/3

2theta = 109.7

theta = 54.7

Hope that helps

Answer 2

Take the square root of both sides.

square root(1/3) = cos theta
1 / square root(3) = cos theta
square root(3) / 3 <-------this is what happens when square root comes out of denominator
square root(3) / 3 = cos theta
arccos/cos^-1 (square root (3) / 3) = theta


There you go and then you just plug that in the calculator and solve.

Answer 3

cos^2 {} = 1+cos 2{} / 2
so...

1/3 = 1+cos 2{} / 2

2/3 = 1 + cos 2{}

-1/3 = cos 2{}

i have to go now.. hopefully this much helps, if not, sorry.

Answer 4

cos^2 @=1/3 so cos @= +-1/3 sqrt3/3
For cos @=1/3sqrt3 we get @=54,74 degrees and @=305.26 degrees
For cos @= -1/3sqrt3 we get @=125.26 deg and @=234.74 deg
So there are four solutions