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

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

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tan^2(x)= 1/3

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

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}

Solve by substitution

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

Let:

tan (x) = u

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

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

4 sin^{2} (x) - 1 = 0

sin (21 0^{\circ})

cos (- \frac{π}{12})

sin (3 x) = 0

sin (\frac{π}{3})

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