What is the general solution for 4tan^2(θ)=12,0<= θ<2pi ?

The general solution for 4tan^2(θ)=12,0<= θ<2pi is θ= pi/3 ,θ=(4pi)/3 ,θ=(2pi)/3 ,θ=(5pi)/3

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4tan^2(θ)=12,0<= θ<2pi

4 tan^{2} (θ) = 12, 0 \leq θ < 2 π

θ = \frac{π}{3}, θ = \frac{4 π}{3}, θ = \frac{2 π}{3}, θ = \frac{5 π}{3}

Degrees

θ = 6 0^{\circ}, θ = 24 0^{\circ}, θ = 12 0^{\circ}, θ = 30 0^{\circ}

Solution steps

4 tan^{2} (θ) = 12, 0 \leq θ < 2 π

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, 0 \leq θ < 2 π : θ = \frac{π}{3}, θ = \frac{4 π}{3}

tan (θ) = 3, 0 \leq θ < 2 π

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{3}, θ = \frac{4 π}{3}

tan (θ) = - 3, 0 \leq θ < 2 π : θ = \frac{2 π}{3}, θ = \frac{5 π}{3}

tan (θ) = - 3, 0 \leq θ < 2 π

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{2 π}{3}, θ = \frac{5 π}{3}

Combine all the solutions

θ = \frac{π}{3}, θ = \frac{4 π}{3}, θ = \frac{2 π}{3}, θ = \frac{5 π}{3}

cos^{2} (x) = 1 - 2 sin^{2} (x)

\frac{1 + tanh ( x )}{1 - tanh ( x )} = 2

2 sin^{2} (θ) + 5 sin (θ) - 3 = 0

sin (2 x) - sin (x) = 0, - π \leq x \leq π

2 (tan (x) + 3) = 5 + tan (x)

What is the general solution for 4tan^2(θ)=12,0<= θ<2pi ?
The general solution for 4tan^2(θ)=12,0<= θ<2pi is θ= pi/3 ,θ=(4pi)/3 ,θ=(2pi)/3 ,θ=(5pi)/3