2sin2 x2cos xSolution2sin2 x2cos x Using the trignometric id

2sin^2 x=2+cos x

Using the trignometric identity : sin^2x +cos^2x =1

we can write the given equation as: 2( 1-cos^2x) = 2+cosx

2 -2cos^^2x = 2+cosx

( 2 gets cancelled on both sides)

cosx + 2cos^2x =0

cosx( 1 +2cosx) =0

Two sets of solution exist: cosx =0 and 1+2cosx =0

Since nothing is given for the solution interval we would find all solutions in the interval : [ 0, 2pi)

cosx =0 ; x= pi/2 , 3pi/2

cosx = -1/2 ; x= pi- pi/3 , pi+pi/3 = 2pi/3 , 4pi/3

Solution set : x = pi/2, 2pi/3 , 3pi/2, 4pi/3