How do you solve #cos^-1 x + cos^-1 2x =60*#?

1 Answer
Oct 1, 2015

#x = +- 1/2#

Explanation:

So from #arccos(x) + arccos(2x) = 60º# take the cosine from both sides

#cos(arccos(x)+arccos(2x)) = 1/2#

Expand the cosine sum, and use #cos(arccos(theta)) = theta# to rewrite it

#x*2x - sin(arccos(x))sin(arccos(2x)) = 1/2#

Use the pythagorean identity #sin^2(x) = 1 - cos^2(x)# to rewrite it. Knowing that the sine is always positive on the arccosine range.

#x*2x - sqrt(1-x^2)*sqrt(1-4x^2)=1/2#
#2x^2 - sqrt((1-x^2)(1-4x^2)) = 1/2#

Isolate the root, then square both sides

#-sqrt((1-x^2)(1-4x^2)) = 1/2 - 2x^2#
#(1-x^2)(1-4x^2) = 1/4 - 2x^2 + 4x^4#

Simplify,

#1 - x^2 -4x^2 + 4x^4 = 1/4 - 2x^2 + 4x^4#

#1 -5x^2 = 1/4 -2x^2#

#1 - 1/4 = 5x^2 -2x^2#

#3/4 = 3x^2#

#1/4 = x^2#

Taking the square root

#x = +- 1/2#