Question #7142d

1 Answer
May 29, 2017

See below.

Explanation:

This is an identity:
cos2x=2cos^2x-1

Plugging it in:
2(2cos^2-1)=cosx+1

4cos^2-2=cosx+1

4cos^2-cosx-3=0

If we set cosx=a, then this is simply a quadratic, with two roots.

4a^2-a-3=0

(4a+3)(a-1)=0

a=-3/4,1 or cosx=-3/4,1

These values all lie within the bounds of cosx, so we can calculate them periodically,

cosx=1 when x=0,2pi,4pi...2npi, or 0,360,720,...360n, where n is an integer.

cosx=-3/4 is not as nice of a number.

cos^-1(-3/4)=2.41885841 radians + 2nπ, or 138.59037789° + 360°n