Question

solve Cos3x+cosx-cos2x=0

Answer 1

Answer:

cos 3x + cos x - cos 2x = 0

[cos 3x + cos x ] - cos 2x = 0

2 cos [(3x + x)/2] .cos [(3x-x)/2] - cos 2x = 0

2 cos 2x . cos x - cos 2x = 0

cos 2x [ 2 cos x - 1 ] = 0

cos 2x = 0 or cos x = 1/2 = cos π/3

2x = (2n +1 )π/2 or x = 2mπ +- π/3

x = (2n + 1 ) π/4 or x = 2mπ +- π/3 ; where n , m ∈ I

Answer 2

Let's solve this,

cos3x+cosx-2cos2x=0

(cos3x+cosx)-2cos2x=0

Using formula for cosx+cosy

cos3x+cosx = 2cos[(3x+x)/2].cos[(3x-x)/2]

=2cos2x.cosx

Putting this in above eqn

2cos2x.cosx-2cos2x=0

cos2x(cosx-1)=0

Solving for general solutions,

For cos2x=0

x=(2n+1)π/4, n∈Z.

For cosx-1=0

x=2nπ, n∈Z.