What is the general solution for solvefor x,cos(x)=1-m^2 ?

The general solution for solvefor x,cos(x)=1-m^2 is x=arccos(1-m^2)+2pin,x=-arccos(1-m^2)+2pin

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solvefor x,cos(x)=1-m^2

solve for

x, cos (x) = 1 - m^{2}

x = arccos (1 - m^{2}) + 2 πn, x = - arccos (1 - m^{2}) + 2 πn

Solution steps

cos (x) = 1 - m^{2}

Apply trig inverse properties

x = arccos (1 - m^{2}) + 2 πn, x = - arccos (1 - m^{2}) + 2 πn

m = - 4 (2.5)

6 x + \frac{1}{3} - 11 x - \frac{2}{9} - \frac{1}{4} (5 x - 2) = \frac{5}{6} (6 x + 1)

m - m \cdot (\frac{- 1}{2}) = \frac{3 - m - 2}{3}

a, 8 a - 5 b = 1

- \frac{22}{15} x + \frac{1}{3} = - \frac{2}{3} + \frac{1}{5}

What is the general solution for solvefor x,cos(x)=1-m^2 ?
The general solution for solvefor x,cos(x)=1-m^2 is x=arccos(1-m^2)+2pin,x=-arccos(1-m^2)+2pin