What is the general solution for sin(z)=-5+j^7 ?

The general solution for sin(z)=-5+j^7 is z=arcsin(-5+j^7)+2pin,z=pi+arcsin(5-j^7)+2pin

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sin(z)=-5+j^7

sin (z) = - 5 + j^{7}

z = arcsin (- 5 + j^{7}) + 2 πn, z = π + arcsin (5 - j^{7}) + 2 πn

Solution steps

sin (z) = - 5 + j^{7}

Apply trig inverse properties

sin (z) = - 5 + j^{7}

General solutions for

sin (z) = - 5 + j^{7}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

z = arcsin (- 5 + j^{7}) + 2 πn, z = π + arcsin (5 - j^{7}) + 2 πn

z = arcsin (- 5 + j^{7}) + 2 πn, z = π + arcsin (5 - j^{7}) + 2 πn

2 cos^{2} (θ) + 5 cos (θ) + 2 = 0

sec^{2} (x) + tan (x) = 0

sin (θ) cot^{2} (θ) - 3 sin (θ) = 0

- 6 sin (x) - 3 = 9

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

What is the general solution for sin(z)=-5+j^7 ?
The general solution for sin(z)=-5+j^7 is z=arcsin(-5+j^7)+2pin,z=pi+arcsin(5-j^7)+2pin