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}
Answer:
3
Step-by-step explanation:
3x - ( x - 4 ) = 3x + 1
3x - x + 4 = 3x + 1
2x + 4 = 3x + 1
3x - 2x = 4 - 1
x = 3
Hope it will help :)
The first step would be to simplify both sides of the equal sign as much as possible.
8x - 5 = -3(3 - 2x)
8x - 5 = -9 - (-6x)
8x - 5 = -9 + 6x
Next, you would need to combine like terms on both sides of the equal sign to one on one side of the equal sign and one on the other.
8x - 5 = -9 + 6x
2x - 5 = -9
2x = -4
Now, all you have to do is isolate the x. To do this, you would divide both sides by 2.
2x = -4
x = -2
I hope this helps!
