How do you solve #cos 2x= cos x-1#?
1 Answer
Jan 29, 2016
Explanation:
Using trig formulae:
# cos 2x = cos^2x - sin^2x = cos^2x - ( 1 - cos^2x ) = 2cos^2x - 1# Replace cos2x by
# (2cos^2 x - 1 )# cos2x = cox - 1 becomes
# 2cos^2x - 1 = cosx - 1 # This is a quadratic function and to solve equate to zero.
hence :
#2cos^2x - 1 - cosx + 1 = 0# simplifies to :
# 2cos^2x - cosx = 0 # factorise : cosx (2cosx - 1 ) = 0
→ cosx = 0 → x =
# 90^@ , 270^@ # and cosx
# = 1/2 →x = 60^@ , 300^@ # These solutions are in the interval 0 < x ≤ 360
