What is the general solution for solvefor x,z=tan(x^3y^2) ?

The general solution for solvefor x,z=tan(x^3y^2) is x=\sqrt[3]{(arctan(z)+pin)/(y^2)}

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solvefor x,z=tan(x^3y^2)

solve for

x, z = tan (x^{3} y^{2})

Solution steps

z = tan (x^{3} y^{2})

Switch sides

tan (x^{3} y^{2}) = z

Apply trig inverse properties

tan (x^{3} y^{2}) = z

General solutions for

tan (x^{3} y^{2}) = z

tan (x) = a \Rightarrow x = arctan (a) + πn

x^{3} y^{2} = arctan (z) + πn

x^{3} y^{2} = arctan (z) + πn

Solve

x^{3} y^{2} = arctan (z) + πn

Divide both sides by

y^{2}; y \neq = 0

x^{3} y^{2} = arctan (z) + πn

Divide both sides by

y^{2}; y \neq = 0

\frac{x ^{3} y ^{2}}{y ^{2}} = \frac{arctan ( z )}{y ^{2}} + \frac{πn}{y ^{2}}; y \neq = 0

Simplify

x^{3} = \frac{arctan ( z )}{y ^{2}} + \frac{πn}{y ^{2}}; y \neq = 0

x^{3} = \frac{arctan ( z )}{y ^{2}} + \frac{πn}{y ^{2}}; y \neq = 0

For

x^{n} = f (a)

, n is odd, the solution is

Simplify

Combine the fractions

\frac{arctan ( z )}{y ^{2}} + \frac{πn}{y ^{2}} : \frac{arctan ( z ) + πn}{y ^{2}}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{arctan ( z ) + πn}{y ^{2}}

sin^{4} (x) = - 2 sin^{2} (x)

3 tan (4 x) = 2

- cos (2 x) = 0

tan (2 x) = - \frac{3}{3}

2 cot^{2} (x) - 3 csc (x) = 0

What is the general solution for solvefor x,z=tan(x^3y^2) ?
The general solution for solvefor x,z=tan(x^3y^2) is x=\sqrt[3]{(arctan(z)+pin)/(y^2)}