What is the general solution for tan^2(θ)-16=0 ?

The general solution for tan^2(θ)-16=0 is θ=1.32581…+pin,θ=-1.32581…+pin

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tan^2(θ)-16=0

tan^{2} (θ) - 16 = 0

θ = 1.32581 \dots + πn, θ = - 1.32581 \dots + πn

Degrees

θ = 75.96375 \dots^{\circ} + 18 0^{\circ} n, θ = - 75.96375 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) - 16 = 0

Solve by substitution

tan^{2} (θ) - 16 = 0

Let:

tan (θ) = u

u^{2} - 16 = 0

u^{2} - 16 = 0 : u = 4, u = - 4

u^{2} - 16 = 0

Move

16

to the right side

u^{2} - 16 = 0

Add

16

to both sides

u^{2} - 16 + 16 = 0 + 16

Simplify

u^{2} = 16

u^{2} = 16

For

x^{2} = f (a)

the solutions are

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

u = 16, u = - 16

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

- 16 = - 4

- 16

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

= - 4

u = 4, u = - 4

Substitute back

u = tan (θ)

tan (θ) = 4, tan (θ) = - 4

tan (θ) = 4, tan (θ) = - 4

tan (θ) = 4 : θ = arctan (4) + πn

tan (θ) = 4

Apply trig inverse properties

tan (θ) = 4

General solutions for

tan (θ) = 4

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

θ = arctan (4) + πn

θ = arctan (4) + πn

tan (θ) = - 4 : θ = arctan (- 4) + πn

tan (θ) = - 4

Apply trig inverse properties

tan (θ) = - 4

General solutions for

tan (θ) = - 4

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

θ = arctan (- 4) + πn

θ = arctan (- 4) + πn

Combine all the solutions

θ = arctan (4) + πn, θ = arctan (- 4) + πn

Show solutions in decimal form

θ = 1.32581 \dots + πn, θ = - 1.32581 \dots + πn

(0.5) (9.8) = (0.45) (9.8) sin (θ)

tan (θ) = - 1.732

sin (\frac{θ}{2}) + cos (θ) = 0

cos (θ) = - 0.75

4 cos (x) = 2 sin (x)

What is the general solution for tan^2(θ)-16=0 ?
The general solution for tan^2(θ)-16=0 is θ=1.32581…+pin,θ=-1.32581…+pin