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Answer 1 · 2017-05-29T20:58:11

See below.

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 #