What is the general solution for arcsin(1/z)=0 ?

The general solution for arcsin(1/z)=0 is No Solution for z\in\mathbb{R}

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arcsin(1/z)=0

arcsin (\frac{1}{z}) = 0

No Solution for z \in R

Solution steps

arcsin (\frac{1}{z}) = 0

Apply trig inverse properties

arcsin (\frac{1}{z}) = 0

arcsin (x) = a \Rightarrow x = sin (a)

\frac{1}{z} = sin (0)

sin (0) = 0

sin (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0

\frac{1}{z} = 0

\frac{1}{z} = 0

Solve

\frac{1}{z} = 0 :

No Solution for

z \in R

\frac{1}{z} = 0

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

1 = 0

The sides are not equal

No Solution for z \in R

No Solution for z \in R

3 sin^{2} (θ) - 4 = - 4 sin (θ)

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

4 cot (θ) sin (θ) = 2

2 sin (2 x) - 5 tan (2 x) = 0

75 = \frac{120}{4} \cdot cos (\frac{2 \cdot π \cdot x}{360}) + \frac{120}{2}

What is the general solution for arcsin(1/z)=0 ?
The general solution for arcsin(1/z)=0 is No Solution for z\in\mathbb{R}