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

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

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cos(z)=0

cos (z) = 0

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

Degrees

z = 9 0^{\circ} + 36 0^{\circ} n, z = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (z) = 0

General solutions for

cos (z) = 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 = \frac{π}{2} + 2 πn, z = \frac{3 π}{2} + 2 πn

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

sin^{2} (x) - 4 sin (x) - 5 = 0

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

sec^{2} (θ) - 4 = 0

2 sin^{2} (x) + sin (x) - 1 = 0, 0 < x < π

sin (x) = - \frac{4}{5}

What is the general solution for cos(z)=0 ?
The general solution for cos(z)=0 is z= pi/2+2pin,z=(3pi)/2+2pin