What is the general solution for tan(θ)(1+tan^2(θ))=0 ?

The general solution for tan(θ)(1+tan^2(θ))=0 is θ=pin

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Popular >

tan(θ)(1+tan^2(θ))=0

Solution

tan (θ) (1 + tan^{2} (θ)) = 0

Solution

θ = πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n

Solution steps

tan (θ) (1 + tan^{2} (θ)) = 0

Solving each part separately

tan (θ) = 0 or 1 + tan^{2} (θ) = 0

tan (θ) = 0 : θ = πn

tan (θ) = 0

General solutions for

tan (θ) = 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}

θ = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

1 + tan^{2} (θ) = 0 :

No Solution

1 + tan^{2} (θ) = 0

Solve by substitution

1 + tan^{2} (θ) = 0

Let:

tan (θ) = u

1 + u^{2} = 0

1 + u^{2} = 0 : u = i, u = - i

1 + u^{2} = 0

Move

1

to the right side

1 + u^{2} = 0

Subtract

1

from both sides

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

Simplify

u^{2} = - 1

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

Substitute back

u = tan (θ)

tan (θ) = i, tan (θ) = - i

tan (θ) = i, tan (θ) = - i

tan (θ) = i :

No Solution

tan (θ) = i

No Solution

tan (θ) = - i :

No Solution

tan (θ) = - i

No Solution

Combine all the solutions

No Solution

Combine all the solutions

θ = πn

Popular Examples

tan(x)=9

tan (x) = 9

sin(2x)=0,0<= x<= 2pi

sin (2 x) = 0, 0 \leq x \leq 2 π

1/2 =sin^2(x)

\frac{1}{2} = sin^{2} (x)

-12sin^2(θ)+3=0

- 12 sin^{2} (θ) + 3 = 0

2cot^2(x)-4=2

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

Frequently Asked Questions (FAQ)

What is the general solution for tan(θ)(1+tan^2(θ))=0 ?
The general solution for tan(θ)(1+tan^2(θ))=0 is θ=pin