What is the general solution for solvefor y,sin(x+y)+sin(y+z)+sin(x+z)=0 ?

The general solution for solvefor y,sin(x+y)+sin(y+z)+sin(x+z)=0 is

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solvefor y,sin(x+y)+sin(y+z)+sin(x+z)=0

solve for

y, sin (x + y) + sin (y + z) + sin (x + z) = 0

y = arcsin (- \frac{sin ( x + z )}{2 cos ( \frac{x - z}{2} )}) + 2 πn - \frac{x}{2} - \frac{z}{2}, y = π + arcsin (\frac{sin ( x + z )}{2 cos ( \frac{x - z}{2} )}) + 2 πn - \frac{x}{2} - \frac{z}{2}

Solution steps

sin (x + y) + sin (y + z) + sin (x + z) = 0

Rewrite using trig identities

sin (x + z) + 2 cos (\frac{x - z}{2}) sin (\frac{x + 2 y + z}{2}) = 0

Move

sin (x + z)

to the right side

2 cos (\frac{x - z}{2}) sin (\frac{x + 2 y + z}{2}) = - sin (x + z)

Divide both sides by

2 cos (\frac{x - z}{2}); x \neq = π + 4 πn + z, x \neq = 3 π + 4 πn + z

sin (\frac{x + 2 y + z}{2}) = - \frac{sin ( x + z )}{2 cos ( \frac{x - z}{2} )}; x \neq = π + 4 πn + z, x \neq = 3 π + 4 πn + z

Apply trig inverse properties

\frac{x + 2 y + z}{2} = arcsin (- \frac{sin ( x + z )}{2 cos ( \frac{x - z}{2} )}) + 2 πn, \frac{x + 2 y + z}{2} = π + arcsin (\frac{sin ( x + z )}{2 cos ( \frac{x - z}{2} )}) + 2 πn

Solve

\frac{x + 2 y + z}{2} = arcsin (- \frac{sin ( x + z )}{2 cos ( \frac{x - z}{2} )}) + 2 πn : y = arcsin (- \frac{sin ( x + z )}{2 cos ( \frac{x - z}{2} )}) + 2 πn - \frac{x}{2} - \frac{z}{2}

Solve

\frac{x + 2 y + z}{2} = π + arcsin (\frac{sin ( x + z )}{2 cos ( \frac{x - z}{2} )}) + 2 πn : y = π + arcsin (\frac{sin ( x + z )}{2 cos ( \frac{x - z}{2} )}) + 2 πn - \frac{x}{2} - \frac{z}{2}

y = arcsin (- \frac{sin ( x + z )}{2 cos ( \frac{x - z}{2} )}) + 2 πn - \frac{x}{2} - \frac{z}{2}, y = π + arcsin (\frac{sin ( x + z )}{2 cos ( \frac{x - z}{2} )}) + 2 πn - \frac{x}{2} - \frac{z}{2}

2 sin (x) - tan (x) = 0

2 sin^{2} (x) - 13 sin (x) + 6 = 0

2 tan (θ) + 3 = 0

tan (x) = \frac{4}{6}

cos (2 x) = - 2 cos^{2} (x)

What is the general solution for solvefor y,sin(x+y)+sin(y+z)+sin(x+z)=0 ?
The general solution for solvefor y,sin(x+y)+sin(y+z)+sin(x+z)=0 is