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

The general solution for d^2+4d+4=sin(x) is x=arcsin(d^2+4d+4)+2pin,x=pi+arcsin(-d^2-4d-4)+2pin

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d^2+4d+4=sin(x)

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

x = arcsin (d^{2} + 4 d + 4) + 2 πn, x = π + arcsin (- d^{2} - 4 d - 4) + 2 πn

Solution steps

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

Switch sides

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

Apply trig inverse properties

x = arcsin (d^{2} + 4 d + 4) + 2 πn, x = π + arcsin (- d^{2} - 4 d - 4) + 2 πn

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sin (6 x) cos (9 x) - cos (6 x) sin (9 x) = - 0.4

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