What is the general solution for sin((2pi)/3-x)+sin((2pi)/3+x)=0 ?

The general solution for sin((2pi)/3-x)+sin((2pi)/3+x)=0 is x=-pi/2-2pin,x=-(3pi)/2-2pin

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sin((2pi)/3-x)+sin((2pi)/3+x)=0

sin (\frac{2 π}{3} - x) + sin (\frac{2 π}{3} + x) = 0

x = - \frac{π}{2} - 2 πn, x = - \frac{3 π}{2} - 2 πn

Solution steps

sin (\frac{2 π}{3} - x) + sin (\frac{2 π}{3} + x) = 0

Rewrite using trig identities

3 cos (- x) = 0

Divide both sides by

3

cos (- x) = 0

General solutions for

cos (- x) = 0

- x = \frac{π}{2} + 2 πn, - x = \frac{3 π}{2} + 2 πn

Solve

- x = \frac{π}{2} + 2 πn : x = - \frac{π}{2} - 2 πn

Solve

- x = \frac{3 π}{2} + 2 πn : x = - \frac{3 π}{2} - 2 πn

x = - \frac{π}{2} - 2 πn, x = - \frac{3 π}{2} - 2 πn

sin (θ) cos (θ) = cot (θ)

cot (θ) = - \frac{4}{3}

tan (x) = \frac{6}{10}

tan (A) = 34

2 sec (x) = 2

What is the general solution for sin((2pi)/3-x)+sin((2pi)/3+x)=0 ?
The general solution for sin((2pi)/3-x)+sin((2pi)/3+x)=0 is x=-pi/2-2pin,x=-(3pi)/2-2pin