What is the general solution for sin^2(x+4)=1 ?

The general solution for sin^2(x+4)=1 is x= pi/2+2pin-4,x=(3pi)/2+2pin-4

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sin^2(x+4)=1

sin^{2} (x + 4) = 1

x = \frac{π}{2} + 2 πn - 4, x = \frac{3 π}{2} + 2 πn - 4

Solution steps

sin^{2} (x + 4) = 1

Solve by substitution

sin (x + 4) = 1, sin (x + 4) = - 1

sin (x + 4) = 1 : x = \frac{π}{2} + 2 πn - 4

sin (x + 4) = - 1 : x = \frac{3 π}{2} + 2 πn - 4

Combine all the solutions

x = \frac{π}{2} + 2 πn - 4, x = \frac{3 π}{2} + 2 πn - 4

\frac{178.9}{sin ( 12 5 ^{\circ} )} = \frac{125}{sin ( x )}

cos (θ) = \frac{3}{8}

arccos (x) - arcsin (x) = arccos (\frac{3}{2})

- 5 tan (θ) - 12 = tan (θ) - 8

\frac{1}{4} sin (4 t) = 0

What is the general solution for sin^2(x+4)=1 ?
The general solution for sin^2(x+4)=1 is x= pi/2+2pin-4,x=(3pi)/2+2pin-4