MATH SOLVE

2 months ago

Q:
# Find all solutions to the equation in the interval [0, 2π).cos 4x - cos 2x = 0

Accepted Solution

A:

solve cos(4x)-cos(2x)=0 ∀ 0<=x<=2pi ..............(0)

Normal solution:

1. use the double angle formula to decompose, and recall cos^2(x)+sin^2(x)=1

cos(4x)=cos^2(2x)-sin^2(2x)=2cos^2(2x)-1 .................(1)

2. substitute (1) in (0)

2cos^2(2x)-1-cos(2x)=0

3. substitute u=cos(2x)

2u^2-u-1=0

4. Solve for x

factor

(u-1)(u+1/2)=0

=> u=1 or u=-1/2

However, since cos(x) is an even function, so solutions to

{cos(2x)=1, cos(-2x)=1, cos(2x)=-1/2 and cos(-2x)} ...........(2)

are all solutions.

5. The cosine function is symmetrical about pi, therefore

cos(-2x)=cos(2*pi-2x),

solution (2) above becomes

{cos(2x)=1, cos(2pi-2x)=1, cos(2x)=-1/2, cos(2pi-2x)=-1/2}

6. Solve each case

cos(2x)=1 => x=0

cos(2pi-2x)=1 => cos(2pi-0)=1 => x=pi

cos(2x)=-1/2 => 2x=2pi/3 or 2x=4pi/3 => x=pi/3 or 2pi/3

cos(2pi-2x)=-1/2 => 2pi-2x=2pi/3 or 2pi-2x=4pi/3 => x=2pi/3 or x=4pi/3

Summing up,

x={0,pi/3, 2pi/3, pi, 4pi/3}

Normal solution:

1. use the double angle formula to decompose, and recall cos^2(x)+sin^2(x)=1

cos(4x)=cos^2(2x)-sin^2(2x)=2cos^2(2x)-1 .................(1)

2. substitute (1) in (0)

2cos^2(2x)-1-cos(2x)=0

3. substitute u=cos(2x)

2u^2-u-1=0

4. Solve for x

factor

(u-1)(u+1/2)=0

=> u=1 or u=-1/2

However, since cos(x) is an even function, so solutions to

{cos(2x)=1, cos(-2x)=1, cos(2x)=-1/2 and cos(-2x)} ...........(2)

are all solutions.

5. The cosine function is symmetrical about pi, therefore

cos(-2x)=cos(2*pi-2x),

solution (2) above becomes

{cos(2x)=1, cos(2pi-2x)=1, cos(2x)=-1/2, cos(2pi-2x)=-1/2}

6. Solve each case

cos(2x)=1 => x=0

cos(2pi-2x)=1 => cos(2pi-0)=1 => x=pi

cos(2x)=-1/2 => 2x=2pi/3 or 2x=4pi/3 => x=pi/3 or 2pi/3

cos(2pi-2x)=-1/2 => 2pi-2x=2pi/3 or 2pi-2x=4pi/3 => x=2pi/3 or x=4pi/3

Summing up,

x={0,pi/3, 2pi/3, pi, 4pi/3}