What is the general solution for solvefor a,cos(a+b)=(sqrt(3))/2 ?

The general solution for solvefor a,cos(a+b)=(sqrt(3))/2 is a= pi/6+2pin-b,a=(11pi)/6+2pin-b

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solvefor a,cos(a+b)=(sqrt(3))/2

solve for

a, cos (a + b) = \frac{3}{2}

a = \frac{π}{6} + 2 πn - b, a = \frac{11 π}{6} + 2 πn - b

Solution steps

cos (a + b) = \frac{3}{2}

General solutions for

cos (a + b) = \frac{3}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a + b = \frac{π}{6} + 2 πn, a + b = \frac{11 π}{6} + 2 πn

a + b = \frac{π}{6} + 2 πn, a + b = \frac{11 π}{6} + 2 πn

Solve

a + b = \frac{π}{6} + 2 πn : a = \frac{π}{6} + 2 πn - b

a + b = \frac{π}{6} + 2 πn

Move

b

to the right side

a + b = \frac{π}{6} + 2 πn

Subtract

b

from both sides

a + b - b = \frac{π}{6} + 2 πn - b

Simplify

a = \frac{π}{6} + 2 πn - b

a = \frac{π}{6} + 2 πn - b