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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lemjay | High School Teacher | (Level 3) Senior Educator

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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`

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