How do you solve #cos(2x)+cos(x)=2# between 0 and 2pi?
1 Answer
Jan 31, 2016
# x = 0 ,2pi #
Explanation:
To solve this type of equation rewrite in terms of cosx
now
#cos2x = cos^2x - sin^2x # and
# sin^2x = 1- cos^2x → cos2x = 2cos^2x - 1 # equation is now :
# 2cos^2x - 1 + cosx = 2 # ( collect terms to left side and equate to zero )
#2 cos^2x + cosx -3 = 0# [To factor : require 2 factors which multiply to -6
and sum to +1 (middle term)] these are -2 and 3 .
hence # (2cosx + 3 )(cosx - 1 ) = 0
so 2cosx + 3 = 0 or cosx - 1 =0
# 2cosx + 3 = 0 → cosx = -3/2 color(black)( " has no solution ") #
# cosx - 1 = 0 → cosx = 1 → x = 0 , 2pi #
