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

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

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1-3tan^2(x)=0

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

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

Solve by substitution

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

Let:

tan (x) = u

1 - 3 u^{2} = 0

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

1 - 3 u^{2} = 0

Move

1

to the right side

1 - 3 u^{2} = 0

Subtract

1

from both sides

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

Simplify

- 3 u^{2} = - 1

- 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

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2 sin (2 x) + 6 sin (x) - 2 cos (x) = 3

sin (2 x) = \frac{4}{5}

2 csc (x) + 17 = 15 + csc (x)

cos (x) - sin (x) = 2

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