How do you solve #2cos2x+3=2# for #0<=x<=360#?

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Answer 1 · 2016-08-28T02:31:57

# x = 60^o, 120^o, 240^o and 300^o#..

Here,

#cos 2x#

#= (2-3)/2#

#=-1/2 #

#=-cos 60^o#

#=cos(180^o-60^o)#

#=cos(120^o#,

and so,

the principal vale of# 2x = 120^o#

The general value is

#(2nX180+-120)^o, n = 0, +-1, +-2, +-3, #, giving,

#x=(nX180+-60)^o, n=0, +-1, +-2, +-3, ...#

For the range #(0^o, 360^o)#, choose + sign and, n = 0, and 1, and

for - sign, n=1 and 2.

Combining and ordering, ,

# x = 60^o, 120^o, 240^o and 300^o#.