How do you find all values of x in the interval [0, 2pi] in the equation #2 sin^2 x = -cos x + 1#?
1 Answer
Jan 31, 2016
# x = 0 ,( 2pi)/3 ,( 4pi)/3 , 2pi#
Explanation:
With this type of equation rewrite in terms of cosines
# sin^2 x = ( 1 - cos^2 x ) # the equation is then :
# 2 (1 - cos^2 x ) = - cosx + 1 # hence
# 2 - 2cos^2 x = - cosx + 1 # collect all terms to left side and equate to zero.
# 2 - 2 cos^2x + cosx - 1 = 0
ie
# 2cos^2x - cosx -1 = 0 # factor this quadratic in the same way as algebraic factoring.
Consider factors of -2 that sum to -1 ( middle term).
(They are -2 + 1 = -1)
hence (2cosx + 1 )(cosx - 1 ) = 0
so >2cox + 1 = 0 or cosx - 1 = 0 →
# cosx = -1/2 , cosx = 1#
# cosx =1 → x = 0 , 2pi #
# cosx = -1/2 → x =( pi - pi/3) =(2pi)/3 , x =( pi + pi/3) = (4pi)/3 #
