If 1 + #cosx# - 2#sin^2x# = 0 , what is the value of #x#?

1 Answer
Jun 11, 2016

Answer:

#x = { pm pi +2k pi} uu {pm pi/3 + 2k pi} # for #k = 0,pm1,pm2,pm3,...#

Explanation:

Making #sin(x)^2=1-cos(x)^2# and #y = cos(x)# we obtain

#1+y-2(1-y^2)=0#

Solving for #y# we have #{y=-1,y=1/2}# so the solutions for #x# must obey

#cos(x) = -1# or #cos(x) = 1/2#

Solving for #x# we obtain

#x = pm pi +2k pi# for #cos(x) = -1# and
#x = pm pi/3 + 2k pi# for #cos(x) = 1/2#
for #k = 0,pm1,pm2,pm3,...#