What is the general solution for solvefor b,a=cos(-1(a/(sqrt(a^2+b^2)))) ?

The general solution for solvefor b,a=cos(-1(a/(sqrt(a^2+b^2)))) is

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solvefor b,a=cos(-1(a/(sqrt(a^2+b^2))))

solve for

b, a = cos (- 1 (\frac{a}{a ^{2} + b ^{2}}))

b = \frac{a ^{2} - a ^{2} arccos ^{2} ( a ) - 4 πn a ^{2} arccos ( a ) - 4 π ^{2} n ^{2} a ^{2}}{arccos ^{2} ( a ) + 4 πn arccos ( a ) + 4 π ^{2} n ^{2}}, b = - \frac{a ^{2} - a ^{2} arccos ^{2} ( a ) - 4 πn a ^{2} arccos ( a ) - 4 π ^{2} n ^{2} a ^{2}}{arccos ^{2} ( a ) + 4 πn arccos ( a ) + 4 π ^{2} n ^{2}}, b = \frac{a ^{2} - a ^{2} arccos ^{2} ( a ) + 4 πn a ^{2} arccos ( a ) - 4 π ^{2} n ^{2} a ^{2}}{arccos ^{2} ( a ) - 4 πn arccos ( a ) + 4 π ^{2} n ^{2}}, b = - \frac{a ^{2} - a ^{2} arccos ^{2} ( a ) + 4 πn a ^{2} arccos ( a ) - 4 π ^{2} n ^{2} a ^{2}}{arccos ^{2} ( a ) - 4 πn arccos ( a ) + 4 π ^{2} n ^{2}}

Solution steps

a = cos (- 1 \cdot (\frac{a}{a ^{2} + b ^{2}}))

Switch sides

cos (- 1 \cdot \frac{a}{a ^{2} + b ^{2}}) = a

Apply trig inverse properties

- 1 \cdot \frac{a}{a ^{2} + b ^{2}} = arccos (a) + 2 πn, - 1 \cdot \frac{a}{a ^{2} + b ^{2}} = - arccos (a) + 2 πn

Solve

- 1 \cdot \frac{a}{a ^{2} + b ^{2}} = arccos (a) + 2 πn : b = \frac{a ^{2} - a ^{2} arccos ^{2} ( a ) - 4 πn a ^{2} arccos ( a ) - 4 π ^{2} n ^{2} a ^{2}}{arccos ^{2} ( a ) + 4 πn arccos ( a ) + 4 π ^{2} n ^{2}}, b = - \frac{a ^{2} - a ^{2} arccos ^{2} ( a ) - 4 πn a ^{2} arccos ( a ) - 4 π ^{2} n ^{2} a ^{2}}{arccos ^{2} ( a ) + 4 πn arccos ( a ) + 4 π ^{2} n ^{2}}

Solve

- 1 \cdot \frac{a}{a ^{2} + b ^{2}} = - arccos (a) + 2 πn : b = \frac{a ^{2} - a ^{2} arccos ^{2} ( a ) + 4 πn a ^{2} arccos ( a ) - 4 π ^{2} n ^{2} a ^{2}}{arccos ^{2} ( a ) - 4 πn arccos ( a ) + 4 π ^{2} n ^{2}}, b = - \frac{a ^{2} - a ^{2} arccos ^{2} ( a ) + 4 πn a ^{2} arccos ( a ) - 4 π ^{2} n ^{2} a ^{2}}{arccos ^{2} ( a ) - 4 πn arccos ( a ) + 4 π ^{2} n ^{2}}

b = \frac{a ^{2} - a ^{2} arccos ^{2} ( a ) - 4 πn a ^{2} arccos ( a ) - 4 π ^{2} n ^{2} a ^{2}}{arccos ^{2} ( a ) + 4 πn arccos ( a ) + 4 π ^{2} n ^{2}}, b = - \frac{a ^{2} - a ^{2} arccos ^{2} ( a ) - 4 πn a ^{2} arccos ( a ) - 4 π ^{2} n ^{2} a ^{2}}{arccos ^{2} ( a ) + 4 πn arccos ( a ) + 4 π ^{2} n ^{2}}, b = \frac{a ^{2} - a ^{2} arccos ^{2} ( a ) + 4 πn a ^{2} arccos ( a ) - 4 π ^{2} n ^{2} a ^{2}}{arccos ^{2} ( a ) - 4 πn arccos ( a ) + 4 π ^{2} n ^{2}}, b = - \frac{a ^{2} - a ^{2} arccos ^{2} ( a ) + 4 πn a ^{2} arccos ( a ) - 4 π ^{2} n ^{2} a ^{2}}{arccos ^{2} ( a ) - 4 πn arccos ( a ) + 4 π ^{2} n ^{2}}

(\frac{1}{2} sin (2 x))^{2} = \frac{1}{4} sin^{2} (4 x)

8 sin^{2} (x) - 3 sin (x) - 2 = 0

2 sin^{2} (x) 3 = 7 sin (x), 0 \leq x \leq 2 π

6 sin (x) + 12 = 15

cot^{2} (x) cos (x) = cos (x)

What is the general solution for solvefor b,a=cos(-1(a/(sqrt(a^2+b^2)))) ?
The general solution for solvefor b,a=cos(-1(a/(sqrt(a^2+b^2)))) is