`sin(x)+sin(3x)=0`

To solve, use the product to sum identities which is:

`sin A + sin B= 2sin((A+B)/2)cos((A-B)/2)`

Applying the identity, the left side of the equation becomes:

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

`2sin((4x)/2)cos((-2x)/2)=0`

`2sin(2x)cos(-x)=0`

To simplify the equation further, apply the double angle identity of sine which is sin(2A)=2sinAcosA.

`2*2sinxcosxcos(-x)=0`

`4sinxcosxcos(-x)=0`

Since cos(-x) = cosx, then the left side becomes:

`4sinxcosxcosx=0`

`4sinxcos^2x=0`

Dividing both sides by 4, the equation simplifies to:

`sinxcos^2x=0`

Then, set each factor equal to zero and solve for x.

For the first factor:

`sinx=0`

`x=0,pi, 2pi`

For the second factor:

`cos^2x=0`

`cosx=0`

`x=pi/2, (3pi)/2`

Hence, `x = 0, pi/2, pi, (3pi)/2` and `2pi` .

Since there is no indicated interval for x, the solution should be express in general form.

**Therefore, the solution to the given equation is**

**`x = pik` **

**and**

**`x=pi/2+pik` **

**where k is any integer.**