What is the general solution for (d^2-2d+3)=sin(x) ?

The general solution for (d^2-2d+3)=sin(x) is x=arcsin(d^2-2d+3)+2pin,x=pi+arcsin(-d^2+2d-3)+2pin

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(d^2-2d+3)=sin(x)

(d^{2} - 2 d + 3) = sin (x)

x = arcsin (d^{2} - 2 d + 3) + 2 πn, x = π + arcsin (- d^{2} + 2 d - 3) + 2 πn

Solution steps

(d^{2} - 2 d + 3) = sin (x)

Switch sides

sin (x) = d^{2} - 2 d + 3

Apply trig inverse properties

sin (x) = d^{2} - 2 d + 3

General solutions for

sin (x) = d^{2} - 2 d + 3

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

x = arcsin (d^{2} - 2 d + 3) + 2 πn, x = π + arcsin (- d^{2} + 2 d - 3) + 2 πn

x = arcsin (d^{2} - 2 d + 3) + 2 πn, x = π + arcsin (- d^{2} + 2 d - 3) + 2 πn

\frac{5}{tan ( x )} = - 2

tan^{22} (x) = sec^{2} (x) - 1

tan^{4} (x) + tan^{2} (x) = tan (x)

sin^{4} (x) + sin^{2} (x) = sin^{6} (x)

cos (a) = \frac{- 11}{14}

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