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

The general solution for d^2+d=cos(x) is x=arccos(d^2+d)+2pin,x=-arccos(d^2+d)+2pin

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d^2+d=cos(x)

d^{2} + d = cos (x)

x = arccos (d^{2} + d) + 2 πn, x = - arccos (d^{2} + d) + 2 πn

Solution steps

d^{2} + d = cos (x)

Switch sides

cos (x) = d^{2} + d

Apply trig inverse properties

x = arccos (d^{2} + d) + 2 πn, x = - arccos (d^{2} + d) + 2 πn

sin (x) = \frac{12}{60}

sin (8 x) = \frac{1}{2}

1 + sin^{2} (x) + cos^{4} (x) = 0

7 \cdot sin^{2} (x) + cos^{2} (x) - 1 = 6 sin (x)

4 sec^{2} (x) - 3 tan^{2} (x) = 5 tan^{2} (x)

What is the general solution for d^2+d=cos(x) ?
The general solution for d^2+d=cos(x) is x=arccos(d^2+d)+2pin,x=-arccos(d^2+d)+2pin