Find the solutions of sin5x + sin3x = 0 in the interval 0, 2pi.

My answer: 0, pi/2, pi, 3pi/2

Looks good to me.

The interval is actually [0, 2pi). Does that change the answer?

To find the solutions of the equation sin(5x) + sin(3x) = 0 in the interval [0, 2π], we can use the trigonometric identity that states sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2).

Therefore, we can rewrite the equation as follows:

2sin((5x+3x)/2)cos((5x-3x)/2) = 0

Simplifying further, we have:

2sin(4x)cos(x) = 0

Now, we know that a product is zero if and only if at least one of the factors is zero. So we can set each factor equal to zero and solve for x.

First, let's solve for sin(4x) = 0:

sin(4x) = 0

Using the unit circle or the trigonometric table, we can determine that sin(4x) = 0 when 4x = 0, π, 2π, 3π, ...

Simplifying, we have:

4x = 0, π, 2π, 3π, ...

Now, dividing by 4 to solve for x, we get:

x = 0, π/4, π/2, 3π/4, ...

Next, let's solve for cos(x) = 0:

cos(x) = 0

Using the unit circle or the trigonometric table, we can determine that cos(x) = 0 when x = π/2 and 3π/2.

Now, combining the solutions for sin(4x) = 0 and cos(x) = 0, we have the following solutions for the equation sin(5x) + sin(3x) = 0 in the interval [0, 2π]:

x = 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4

So the solutions in the given interval are: 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4.