What is the general solution for cos(z+1)=0 ?

The general solution for cos(z+1)=0 is z= pi/2+2pin-1,z=(3pi)/2+2pin-1

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cos(z+1)=0

cos (z + 1) = 0

z = \frac{π}{2} + 2 πn - 1, z = \frac{3 π}{2} + 2 πn - 1

Degrees

z = 32.70422 \dots^{\circ} + 36 0^{\circ} n, z = 212.70422 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos (z + 1) = 0

General solutions for

cos (z + 1) = 0

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}

z + 1 = \frac{π}{2} + 2 πn, z + 1 = \frac{3 π}{2} + 2 πn

z + 1 = \frac{π}{2} + 2 πn, z + 1 = \frac{3 π}{2} + 2 πn

Solve

z + 1 = \frac{π}{2} + 2 πn : z = \frac{π}{2} + 2 πn - 1

z + 1 = \frac{π}{2} + 2 πn

Move

1

to the right side

z + 1 = \frac{π}{2} + 2 πn

Subtract

1

from both sides

z + 1 - 1 = \frac{π}{2} + 2 πn - 1

Simplify

z = \frac{π}{2} + 2 πn - 1

z = \frac{π}{2} + 2 πn - 1