What is the general solution for solvefor a,sin(a+b/2)=cos(c/2) ?

The general solution for solvefor a,sin(a+b/2)=cos(c/2) is a=(pi-c-b)/2+2pin,a=pi+2pin+(-pi+c-b)/2

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solvefor a,sin(a+b/2)=cos(c/2)

solve for

a, sin (a + \frac{b}{2}) = cos (\frac{c}{2})

a = \frac{π - c - b}{2} + 2 πn, a = π + 2 πn + \frac{- π + c - b}{2}

Solution steps

sin (a + \frac{b}{2}) = cos (\frac{c}{2})

Rewrite using trig identities

sin (a + \frac{b}{2}) = sin (\frac{π}{2} - \frac{c}{2})

Apply trig inverse properties

a + \frac{b}{2} = \frac{π}{2} - \frac{c}{2} + 2 πn, a + \frac{b}{2} = π - (\frac{π}{2} - \frac{c}{2}) + 2 πn

a + \frac{b}{2} = \frac{π}{2} - \frac{c}{2} + 2 πn : a = \frac{π - c - b}{2} + 2 πn

a + \frac{b}{2} = π - (\frac{π}{2} - \frac{c}{2}) + 2 πn : a = π + 2 πn + \frac{- π + c - b}{2}

a = \frac{π - c - b}{2} + 2 πn, a = π + 2 πn + \frac{- π + c - b}{2}

tan (x) - cot (x) = 2 cot^{2} (x)

cot (\frac{- x + 3 n}{4}) = 0

sin (x) + sin^{2} (x) + sin^{3} (x) = 1

\frac{cos ( a )}{1 + sin ( a )} = sec (a)

cos^{2} (x) + sin^{2} (2 x) = 0

What is the general solution for solvefor a,sin(a+b/2)=cos(c/2) ?
The general solution for solvefor a,sin(a+b/2)=cos(c/2) is a=(pi-c-b)/2+2pin,a=pi+2pin+(-pi+c-b)/2