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

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

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tan^2(x)-4=0

tan^{2} (x) - 4 = 0

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

Degrees

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

Solution steps

tan^{2} (x) - 4 = 0

Solve by substitution

tan^{2} (x) - 4 = 0

Let:

tan (x) = 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 (x)

tan (x) = 2, tan (x) = - 2

tan (x) = 2, tan (x) = - 2

tan (x) = 2 : x = arctan (2) + πn

tan (x) = 2

Apply trig inverse properties

tan (x) = 2

General solutions for

tan (x) = 2

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

x = arctan (2) + πn

x = arctan (2) + πn

tan (x) = - 2 : x = arctan (- 2) + πn

tan (x) = - 2

Apply trig inverse properties

tan (x) = - 2

General solutions for

tan (x) = - 2

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

x = arctan (- 2) + πn

x = arctan (- 2) + πn

Combine all the solutions

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

Show solutions in decimal form

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

2 sin^{2} (θ) + sin (θ) - 1 = 0, 0 \leq θ < 2 π

2 cos^{2} (x) - 3 cos (x) = 2

cot (θ) = \frac{1}{3}

- 3 sin^{2} (θ) + 4 sin (θ) - 7 = - 6

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

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