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

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

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3tan^2(x)=1

3 tan^{2} (x) = 1

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

3 tan^{2} (x) = 1

Solve by substitution

3 tan^{2} (x) = 1

Let:

tan (x) = u

3 u^{2} = 1

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

3 u^{2} = 1

Divide both sides by

3

3 u^{2} = 1

Divide both sides by

3

\frac{3 u ^{2}}{3} = \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 (2 θ) = \frac{1}{2}

cos (\frac{3 π}{2})

sin (θ) = - \frac{3}{2}

arcsec (- 2)

cos^{2} (x) - 1 = 0

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