How to find the trigonometric equation for #3(cos x - 2sin^2 x) = cos x - 2# ?

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Answer 1 · 2018-08-07T02:23:13

#2 cos x - 3 cos 2x = 1#

#3(cos x - 2 sin^2 x) = cos x - 2#

#1 - 2sin^2 x = cos 2x#

#:. 3 (cos x - cos 2x -1) = cos x - 2#

#3 cos x - 3 cos 2x - 3 = cos x - 2#

#3 cos x - cos x - 3 cos 2x = 3 - 2#

#2 cos x - 3 cos 2x = 1#

Is this your requirement?

Answer 2 · 2018-08-07T05:06:30

# (cos x + 1 )(cos x - 2/3 ) = 0 #
#rArr x = { ( 2 k + 1 ) pi}, { 2kpi +- cos^(-1)( 2/3 )}#,
#k = 0, +-1, +-2, +-3, ...# -

Using #sin^2x = 1- cos^2x and c = cos x#,

#6c^2 + 2 c - 4 = 2 ( 3 c^2 + c - 2 ) = 0#

#rArr (cos x+1)( cos x - 2/3 ) = 0