Find all solutions in the interval [0,2π)
sin3x−sinx=0
Identity:
sin 3θ = 3 sin θ -4 sin^3 θ
so
3 sin x - 4 sin^3 x = sin x
Immediately this is true when x = 0
where else?
3 - 4 sin^2 x = 1
2 = 4 sin^2 x
sin^2 x = 1/2
sin x = + or - 1/sqrt2
LOL try 45 degrees, 135 degrees, 225 and 315 (convert those to radians )
To find all the solutions of the equation sin(3x) - sin(x) = 0 in the interval [0, 2π), we need to solve the equation for x.
Let's start by applying the identity for the difference of two sines:
sin(a) - sin(b) = 2 * cos((a + b) / 2) * sin((a - b) / 2)
Using this identity, we rewrite the equation as:
2 * cos((3x + x) / 2) * sin((3x - x) / 2) = 0
Simplifying further:
2 * cos(2x) * sin(x) = 0
Now, we have two factors that can make the equation equal to zero: cos(2x) = 0 and sin(x) = 0.
1. cos(2x) = 0:
To find the solutions for cos(2x) = 0, we solve for x in the equation:
2x = π/2 + kπ, where k is an integer
Dividing both sides by 2, we get:
x = π/4 + kπ/2, where k is an integer
Note that in the given interval [0, 2π), we need to consider the values of k for which the solutions are between 0 and 2π. The possible values of k are 0, 1, 2, and 3.
For k = 0: x = π/4
For k = 1: x = 3π/4
For k = 2: x = 5π/4
For k = 3: x = 7π/4
2. sin(x) = 0:
To find the solutions for sin(x) = 0, we solve for x in the equation:
x = kπ, where k is an integer
Again, in the given interval [0, 2π), we need to consider the values of k for which the solutions are between 0 and 2π. The possible values of k are 0, 1, and 2.
For k = 0: x = 0
For k = 1: x = π
For k = 2: x = 2π
Combining the solutions from both cases, the solutions for the equation sin(3x) - sin(x) = 0 in the interval [0, 2π) are:
x = 0, π/4, π, 3π/4, 5π/4, 7π/4, 2π