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

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

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3tan^2(θ)-1=0

3 tan^{2} (θ) - 1 = 0

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

Degrees

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

Solution steps

3 tan^{2} (θ) - 1 = 0

Solve by substitution

3 tan^{2} (θ) - 1 = 0

Let:

tan (θ) = u

3 u^{2} - 1 = 0

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

3 u^{2} - 1 = 0

Move

1

to the right side

3 u^{2} - 1 = 0

Add

1

to both sides

3 u^{2} - 1 + 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 (θ)

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

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

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

tan (θ) = \frac{1}{3}

Apply trig inverse properties

tan (θ) = \frac{1}{3}

General solutions for

tan (θ) = \frac{1}{3}

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

θ = arctan (\frac{1}{3}) + πn

θ = arctan (\frac{1}{3}) + πn

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

tan (θ) = - \frac{1}{3}

Apply trig inverse properties

tan (θ) = - \frac{1}{3}

General solutions for

tan (θ) = - \frac{1}{3}

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

θ = arctan (- \frac{1}{3}) + πn

θ = arctan (- \frac{1}{3}) + πn

Combine all the solutions

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

Show solutions in decimal form

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

sin (2 θ - \frac{π}{2}) = - 1

sin (\frac{x}{2}) + cos (x) = 0

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

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

sin^{2} (x) - 2 sin (x) - 3 = 0

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