What is the general solution for tan^2(θ)=3,0<= θ<= 2pi ?

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

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tan^2(θ)=3,0<= θ<= 2pi

tan^{2} (θ) = 3, 0 \leq θ \leq 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

tan^{2} (θ) = 3, 0 \leq θ \leq 2 π

Solve by substitution

tan^{2} (θ) = 3

Let:

tan (θ) = u

u^{2} = 3

u^{2} = 3 : u = 3, u = - 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 \leq 2 π : θ = \frac{π}{3}, θ = \frac{4 π}{3}

tan (θ) = 3, 0 \leq θ \leq 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 \leq 2 π

θ = \frac{π}{3}, θ = \frac{4 π}{3}

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

tan (θ) = - 3, 0 \leq θ \leq 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 \leq 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}

3 sin (x) - 2 cos^{2} (x) = 0

4 cos^{2} (x) + 4 cos (x) - 3 = 0

sin (4 x) cos (x) - cos (4 x) sin (x) = 0

2 sin (x) - (2 - 2) = 2 csc (x)

5 sin (62.83 x) = - 0.32

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