What is the general solution for arctan(1+x^2)=0 ?

The general solution for arctan(1+x^2)=0 is No Solution for x\in\mathbb{R}

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arctan(1+x^2)=0

arctan (1 + x^{2}) = 0

No Solution for x \in R

Solution steps

arctan (1 + x^{2}) = 0

Apply trig inverse properties

arctan (1 + x^{2}) = 0

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

1 + x^{2} = 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

1 + x^{2} = 0

1 + x^{2} = 0

Solve

1 + x^{2} = 0 :

No Solution for

x \in R

1 + x^{2} = 0

Move

1

to the right side

1 + x^{2} = 0

Subtract

1

from both sides

1 + x^{2} - 1 = 0 - 1

Simplify

x^{2} = - 1

x^{2} = - 1

x^{2}

cannot be negative for

x \in R

No Solution for x \in R

No Solution for x \in R

θ, tan (θ) = - 1

2 sin (4 x) = - 4 sin (2 x)

cos (x) = (\frac{5}{3}), cot (2 x)

\frac{sec ( x )}{tan ( x )} = \frac{cos ( x )}{( sin ^{2} ( x ) )}

x, tan (x) = \frac{1}{8}

What is the general solution for arctan(1+x^2)=0 ?
The general solution for arctan(1+x^2)=0 is No Solution for x\in\mathbb{R}