Question

cosx +cos3x-2cos2x=0

Answer 1

Answer:

x = nπ ± π / 4 or x = 2nπ

Step-by-step explanation:

Given---> Cosx + Cos3x - 2Cos2x = 0

To find ---> General value of x

Solution---> General value of Cos is

α = 2nπ ± θ

ATQ,

Cosx + Cos3x - 2Cos2x = 0

=> Cos3x + Cosx - 2Cos2x = 0

We know that,

CosC + CosD = 2Cos (C+D/2) Cos (C-D/2)

Applying it we get ,

=> 2 Cos {(3x+x )/2} Cos {(3x - x)/2} - 2Cos2x = 0

=> 2 Cos ( 4x / 2 ) Cos ( 2x / 2 ) - 2 Cos2x = 0

=> 2 Cos2x Cosx - 2 Cos2x = 0

=> 2 Cos2x ( Cosx - 1 ) = 0

If Cos2x = 0

=> Cos2x = Cos π/2

=> Cos2x = Cos ( 2nπ ± π / 2 )

=> 2x = 2nπ ± π /2

=> x = 2nπ / 2 ± π / 4

=> x = nπ ± π / 4

If Cosx - 1 = 0

=> Cosx = 1

=> Cosx = Cos0°

=> Cosx = Cos ( 2nπ ± 0 )

=> x = 2nπ

Answer 2

Answer:

Step-by-step explanation:

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.