Find the general solution of
sin x + sin 2x + sin 3x + sin 4x = 0 ?
since you will wind up with a 4th-degree polynomial in sinx, it's gonna take some doing. Take a peek at wolframalpha.com
To find the general solution of the equation sin x + sin 2x + sin 3x + sin 4x = 0, we can start by using the trigonometric identity:
sin A + sin B = 2sin((A+B)/2)cos((A-B)/2)
Applying this identity to the equation, we have:
sin x + sin 2x + sin 3x + sin 4x = 0
2sin(5x/2)cos(-x/2) + 2sin(5x/2)cos(x/2) = 0
2sin(5x/2)(cos(x/2) - cos(-x/2)) = 0
2sin(5x/2)(2sin(x/2)sin(-x/2)) = 0
2sin(5x/2)(-2sin^2(x/2)) = 0
-4sin^2(x/2)sin(5x/2) = 0
Now we have two cases to consider:
Case 1: -4sin^2(x/2) = 0
This means that sin^2(x/2) = 0. Since sin^2(x/2) = (1-cos(x))/2, we can set 1 - cos(x) = 0, which gives cos(x) = 1. Therefore, x = 2πn, where n is an integer.
Case 2: sin(5x/2) = 0
This means that 5x/2 = kπ, where k is an integer. Solving for x gives x = 2kπ/5.
Combining both cases, the general solution to the equation sin x + sin 2x + sin 3x + sin 4x = 0 is:
x = 2πn, 2kπ/5
where n and k are integers.