What is the general solution for solvefor C,arctan((1/(2*pi*R*C))/f)=0 ?

The general solution for solvefor C,arctan((1/(2*pi*R*C))/f)=0 is No Solution for C\in\mathbb{R}

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solvefor C,arctan((1/(2piR*C))/f)=0

solve for

C, arctan (\frac{\frac{1}{2 \cdot π \cdot R \cdot C}}{f}) = 0

No Solution for C \in R

Solution steps

arctan (\frac{\frac{1}{2 π RC}}{f}) = 0

Apply trig inverse properties

arctan (\frac{\frac{1}{2 π RC}}{f}) = 0

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

\frac{\frac{1}{2 π RC}}{f} = tan (0)

tan (0) = 0

tan (0)

Use the following trivial identity:

tan (0) = 0

tan (0)

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 0

= 0

\frac{\frac{1}{2 π RC}}{f} = 0

\frac{\frac{1}{2 π RC}}{f} = 0

Solve

\frac{\frac{1}{2 π RC}}{f} = 0 :

No Solution for

C \in R

\frac{\frac{1}{2 π RC}}{f} = 0

Simplify

\frac{\frac{1}{2 π RC}}{f} : \frac{1}{2 π CR f}

\frac{\frac{1}{2 π RC}}{f}

Apply the fraction rule:

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

= \frac{1}{2 π RC f}

\frac{1}{2 π CR f} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 = 0

The sides are not equal

No Solution for C \in R

No Solution for C \in R

sin (x) = - 2 cos (x)

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

sin (x) - cos (2 x) - 1 = 0

sin^{2} (x) = 3

x, c^{2} = a^{2} + b^{2} - 2 ab \cdot cos (x)

What is the general solution for solvefor C,arctan((1/(2*pi*R*C))/f)=0 ?
The general solution for solvefor C,arctan((1/(2*pi*R*C))/f)=0 is No Solution for C\in\mathbb{R}