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

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

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tan^2(x)=3,0<x<2pi

tan^{2} (x) = 3, 0 < x < 2 π

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

Degrees

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

Solution steps

tan^{2} (x) = 3, 0 < x < 2 π

Solve by substitution

tan^{2} (x) = 3

Let:

tan (x) = 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 (x)

tan (x) = 3, tan (x) = - 3

tan (x) = 3, tan (x) = - 3

tan (x) = 3, 0 < x < 2 π : x = \frac{π}{3}, x = \frac{4 π}{3}

tan (x) = 3, 0 < x < 2 π

General solutions for

tan (x) = 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}

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

Solutions for the range

0 < x < 2 π

x = \frac{π}{3}, x = \frac{4 π}{3}

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

tan (x) = - 3, 0 < x < 2 π

General solutions for

tan (x) = - 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}

x = \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

Solutions for the range

0 < x < 2 π

x = \frac{2 π}{3}, x = \frac{5 π}{3}

Combine all the solutions

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

4 cos (x - π) - 4 = 0

cos (θ) = \frac{4 2}{7}, sin (θ)

arctan (\frac{1}{x - 1}) + arctan (\frac{2}{x + 1}) = \frac{π}{4}

cot (3 x) = - tan (- \frac{2 π}{5})

2 - 2 sin (x) = 3

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