What is the general solution for tan^2(x)-1/3 =0 ?

The general solution for tan^2(x)-1/3 =0 is x=0.52359…+pin,x=-0.52359…+pin

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tan^2(x)-1/3 =0

tan^{2} (x) - \frac{1}{3} = 0

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

Degrees

x = 3 0^{\circ} + 18 0^{\circ} n, x = - 3 0^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) - \frac{1}{3} = 0

Solve by substitution

tan^{2} (x) - \frac{1}{3} = 0

Let:

tan (x) = u

u^{2} - \frac{1}{3} = 0

u^{2} - \frac{1}{3} = 0 : u = \frac{1}{3}, u = - \frac{1}{3}

u^{2} - \frac{1}{3} = 0

Move

\frac{1}{3}

to the right side

u^{2} - \frac{1}{3} = 0

Add

\frac{1}{3}

to both sides

u^{2} - \frac{1}{3} + \frac{1}{3} = 0 + \frac{1}{3}

Simplify

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{3}, u = - \frac{1}{3}

Substitute back

u = tan (x)

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{3}

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{3}

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

tan (x) = \frac{1}{3}

Apply trig inverse properties

tan (x) = \frac{1}{3}

General solutions for

tan (x) = \frac{1}{3}

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

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

tan (x) = - \frac{1}{3} : x = arctan (- \frac{1}{3}) + πn

tan (x) = - \frac{1}{3}

Apply trig inverse properties

tan (x) = - \frac{1}{3}

General solutions for

tan (x) = - \frac{1}{3}

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

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

Combine all the solutions

x = arctan (\frac{1}{3}) + πn, x = arctan (- \frac{1}{3}) + πn

Show solutions in decimal form

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

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

sin^{2} (x) - 1 = 0, 0 \leq x < 2 π

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

sin^{4} (x) - cos^{4} (x) = 0

tan (θ) cos (θ) + cos (θ) = 0

What is the general solution for tan^2(x)-1/3 =0 ?
The general solution for tan^2(x)-1/3 =0 is x=0.52359…+pin,x=-0.52359…+pin