What is the general solution for 4tan^2(θ)=12 ?

The general solution for 4tan^2(θ)=12 is θ= pi/3+pin,θ=(2pi)/3+pin

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4tan^2(θ)=12

4 tan^{2} (θ) = 12

θ = \frac{π}{3} + πn, θ = \frac{2 π}{3} + πn

Degrees

θ = 6 0^{\circ} + 18 0^{\circ} n, θ = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

4 tan^{2} (θ) = 12

Solve by substitution

4 tan^{2} (θ) = 12

Let:

tan (θ) = u

4 u^{2} = 12

4 u^{2} = 12 : u = 3, u = - 3

4 u^{2} = 12

Divide both sides by

4

4 u^{2} = 12

Divide both sides by

4

\frac{4 u ^{2}}{4} = \frac{12}{4}

Simplify

u^{2} = 3

u^{2} = 3

For

x^{2} = f (a)

the solutions are

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

u = 3, u = - 3

Substitute back

u = tan (θ)

tan (θ) = 3, tan (θ) = - 3

tan (θ) = 3, tan (θ) = - 3

tan (θ) = 3 : θ = \frac{π}{3} + πn

tan (θ) = 3

General solutions for

tan (θ) = 3

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 = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

tan (θ) = - 3 : θ = \frac{2 π}{3} + πn

tan (θ) = - 3

General solutions for

tan (θ) = - 3

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 = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

Combine all the solutions

θ = \frac{π}{3} + πn, θ = \frac{2 π}{3} + πn

4 sin (x) = 3

0.5 = cos (x)

sin (3 x) - sin (6 x) = 0

2 sin (θ) + cos (θ) = 0

2 tan (x) sin (x) - 3 sin (x) = 0

What is the general solution for 4tan^2(θ)=12 ?
The general solution for 4tan^2(θ)=12 is θ= pi/3+pin,θ=(2pi)/3+pin