What is the general solution for 6tan^2(x)-2=0 ?

The general solution for 6tan^2(x)-2=0 is x=0.52359…+pin,x=-0.52359…+pin

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6tan^2(x)-2=0

6 tan^{2} (x) - 2 = 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

6 tan^{2} (x) - 2 = 0

Solve by substitution

6 tan^{2} (x) - 2 = 0

Let:

tan (x) = u

6 u^{2} - 2 = 0

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

6 u^{2} - 2 = 0

Move

2

to the right side

6 u^{2} - 2 = 0

Add

2

to both sides

6 u^{2} - 2 + 2 = 0 + 2

Simplify

6 u^{2} = 2

6 u^{2} = 2

Divide both sides by

6

6 u^{2} = 2

Divide both sides by

6

\frac{6 u ^{2}}{6} = \frac{2}{6}

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

tan (x) = \infty

2 arctanh (\frac{( x - 2 )}{( x + 1 )}) = ln (2)

tan (x) = - 1.5

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

5 sin (θ) tan (θ) - 10 tan (θ) + 3 sin (θ) - 6 = 0

What is the general solution for 6tan^2(x)-2=0 ?
The general solution for 6tan^2(x)-2=0 is x=0.52359…+pin,x=-0.52359…+pin