What is the general solution for tan^2(θ)-4=0 ?

The general solution for tan^2(θ)-4=0 is θ=1.10714…+pin,θ=-1.10714…+pin

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tan^2(θ)-4=0

tan^{2} (θ) - 4 = 0

θ = 1.10714 \dots + πn, θ = - 1.10714 \dots + πn

Degrees

θ = 63.43494 \dots^{\circ} + 18 0^{\circ} n, θ = - 63.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) - 4 = 0

Solve by substitution

tan^{2} (θ) - 4 = 0

Let:

tan (θ) = u

u^{2} - 4 = 0

u^{2} - 4 = 0 : u = 2, u = - 2

u^{2} - 4 = 0

Move

4

to the right side

u^{2} - 4 = 0

Add

4

to both sides

u^{2} - 4 + 4 = 0 + 4

Simplify

u^{2} = 4

u^{2} = 4

For

x^{2} = f (a)

the solutions are

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

u = 4, u = - 4

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

- 4 = - 2

- 4

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= - 2

u = 2, u = - 2

Substitute back

u = tan (θ)

tan (θ) = 2, tan (θ) = - 2

tan (θ) = 2, tan (θ) = - 2

tan (θ) = 2 : θ = arctan (2) + πn

tan (θ) = 2

Apply trig inverse properties

tan (θ) = 2

General solutions for

tan (θ) = 2

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

θ = arctan (2) + πn

θ = arctan (2) + πn

tan (θ) = - 2 : θ = arctan (- 2) + πn

tan (θ) = - 2

Apply trig inverse properties

tan (θ) = - 2

General solutions for

tan (θ) = - 2

tan (x) = - a \Rightarrow x = arctan (- a) + πn

θ = arctan (- 2) + πn

θ = arctan (- 2) + πn

Combine all the solutions

θ = arctan (2) + πn, θ = arctan (- 2) + πn

Show solutions in decimal form

θ = 1.10714 \dots + πn, θ = - 1.10714 \dots + πn

csc^{2} (θ) - 2 csc (θ) - 8 = 0

(tan (θ) + 1) (cos (θ) + 1) = 0

sec (θ) = \frac{5}{3}

1 - cos (x) = 3 sin (x)

sin (θ) = \frac{7}{25}

What is the general solution for tan^2(θ)-4=0 ?
The general solution for tan^2(θ)-4=0 is θ=1.10714…+pin,θ=-1.10714…+pin