What is the general solution for solvefor x,z=arctan(y/x) ?

The general solution for solvefor x,z=arctan(y/x) is x= y/(tan(z))

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solvefor x,z=arctan(y/x)

solve for

x, z = arctan (\frac{y}{x})

x = \frac{y}{tan ( z )}

Solution steps

z = arctan (\frac{y}{x})

Switch sides

arctan (\frac{y}{x}) = z

Apply trig inverse properties

arctan (\frac{y}{x}) = z

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

\frac{y}{x} = tan (z)

\frac{y}{x} = tan (z)

Solve

\frac{y}{x} = tan (z) : x = \frac{y}{tan ( z )}; z \neq = πn

\frac{y}{x} = tan (z)

Multiply both sides by

x

\frac{y}{x} = tan (z)

Multiply both sides by

x

\frac{y}{x} x = tan (z) x

Simplify

y = tan (z) x

y = tan (z) x

Switch sides

tan (z) x = y

Divide both sides by

tan (z); z \neq = πn

tan (z) x = y

Divide both sides by

tan (z); z \neq = πn

\frac{tan ( z ) x}{tan ( z )} = \frac{y}{tan ( z )}; z \neq = πn

Simplify

x = \frac{y}{tan ( z )}; z \neq = πn

x = \frac{y}{tan ( z )}; z \neq = πn

x = \frac{y}{tan ( z )}

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

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

cos^{2} (2 x + \frac{π}{6}) = \frac{1}{2}

cos (5 x) = 1, 0 \leq x \leq 2 π

sec (2 θ) = 2, 0 < θ < \frac{π}{2}

What is the general solution for solvefor x,z=arctan(y/x) ?
The general solution for solvefor x,z=arctan(y/x) is x= y/(tan(z))