SOLUTION: Good morning, can somebody help me factorize and solve step by step following equation?: sin(2x)cos(pi/3) - cos(2x)sin(pi/3) + 3^(1/2)(cos(2x)cos(pi/3) + sin(2x)sin(pi/3))-3^(1/2)

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Question 1062122: Good morning, can somebody help me factorize and solve step by step following equation?:
sin(2x)cos(pi/3) - cos(2x)sin(pi/3) + 3^(1/2)(cos(2x)cos(pi/3) + sin(2x)sin(pi/3))-3^(1/2) = 0
Thank you very much RB
Found 2 solutions by rothauserc, Edwin McCravy:
Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website!
-sqrt(3) - 1/2 sqrt(3) cos(2 x) + 1/2 sin(2 x) + sqrt(3) (1/2 cos(2 x) + 1/2 sqrt(3) sin(2 x)) = 0
:
Simplify and substitute y = 1/2 sin(2 x)
:
-sqrt(3) - 1/2 sqrt(3) cos(2 x) + 1/2 sin(2 x) + sqrt(3) (1/2 cos(2 x) + 1/2 sqrt(3) sin(2 x)) = (4 sin(2 x))/2 - sqrt(3) = 4y - sqrt(3) = 0
:
4y - sqrt(3) = 0
:
Add sqrt(3) to both sides
:
4y = sqrt(3)
:
Divide both sides by 4
:
y = sqrt(3)/4
:
Substitute back for y = 1/2 sin(2 x)
:
1/2 sin(2x) = sqrt(3)/4
:
Multiply both sides by 2
:
sin(2x) = sqrt(3)/2
:
Take the inverse sine of both sides
:
2x = (2π)/3 + 2πn1 for n1 in Z
or 2x = π/3 + 2πn2 for n2 in Z
:
Simplify each equation
:
Divide both sides by 2
:
x = π/3 + πn1 for n1 in Z
or 2x = π/3 + 2πn2 for n2 in Z
:
Divide both sides by 2
:
*************************************
x = π/3 + πn1 for n1 in Z
:
or x = π/6 + πn2 for n2 in Z
*************************************
:

Answer by Edwin McCravy(20060) (Show Source):
You can put this solution on YOUR website!

Since and The 2nd and 3rd terms cancel out, and the 1st and 4th terms are like terms and can be combined Edwin