# Using the sum-to-product formula, find the solution: `sin(x) + sin(3x) = sin(2x)`I reach `2sin(2x)cos(x)=sin(2x)` If this is correct, how do I separate the `cos(x)` to solve?

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### 1 Answer

Yes, your solution is correct.

From there, divide both sides by `sin(2x` ).

`[2sin(2x)cos(x)]/sin(2x) = [sin(2x)]/[sin(2x)]`

`2cos (x) = 1`

` cos (x) = 1/2`

Then, refer to Unit Circle Chart to determine the value of angle x.

` x = pi/3 and (5pi)/3`

Hence, the general solution is:

`x_1 = pi/3 + 2pik ` and `x_2 = (5pi)/3 + 2pik`

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To verify, let's go over the solution.

`sin(x) + sinx (3x) = sin (2x)`

Express 3x as sum of two angles.

`sin(x) + sin(x + 2x) = sin (2x)`

Use the formula for sum of angles of sine. `sin(A+B) = sinAcosB + cosBsinA`.

`sin (x) + sin (x) cos(2x) + cos (x) sin (2x) = sin (2x)`

Use the double angle identity of cosine. `cos 2A = 2cos^2 - 1`

`sin(x) + sin(x)[2cos^2 (x) - 1] + cos(x)sin(2x) = sin(2x)`

`sin(x) + sin(x)[2cos^2(x)-1] + cos(x)sin(2x)= sin (2x)`

`sin(x) + 2sin(x)cos^2(x) - sin(x) + cos(x)sin(2x) = sin(2x)`

`2sin(x) cos^2(x) + cos(x)sin(2x) = sin(2x)`

`cos(x)sin(2x) + cos(x) sin(2x) = sin(2x)`

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

Divide both sides by sin (2x).

`2cos (x) = 1`

`cos (x) = 1/2`

`x = pi/3 and (5pi)/3 `

**Answer:** `x_1 = pi/3 + 2pik` **and** `x_2 = (5pi)/3 + 2pik`