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

The general solution for 3cot^2(θ)-1=0 is θ=1.04719…+pin,θ=2.09439…+pin

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3cot^2(θ)-1=0

3 cot^{2} (θ) - 1 = 0

θ = 1.04719 \dots + πn, θ = 2.09439 \dots + πn

Degrees

θ = 6 0^{\circ} + 18 0^{\circ} n, θ = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

3 cot^{2} (θ) - 1 = 0

Solve by substitution

3 cot^{2} (θ) - 1 = 0

Let:

cot (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 = cot (θ)

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

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

cot (θ) = \frac{1}{3} : θ = arccot (\frac{1}{3}) + πn

cot (θ) = \frac{1}{3}

Apply trig inverse properties

cot (θ) = \frac{1}{3}

General solutions for

cot (θ) = \frac{1}{3}

cot (x) = a \Rightarrow x = arccot (a) + πn

θ = arccot (\frac{1}{3}) + πn

θ = arccot (\frac{1}{3}) + πn

cot (θ) = - \frac{1}{3} : θ = arccot (- \frac{1}{3}) + πn

cot (θ) = - \frac{1}{3}

Apply trig inverse properties

cot (θ) = - \frac{1}{3}

General solutions for

cot (θ) = - \frac{1}{3}

cot (x) = - a \Rightarrow x = arccot (- a) + πn

θ = arccot (- \frac{1}{3}) + πn

θ = arccot (- \frac{1}{3}) + πn

Combine all the solutions

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

Show solutions in decimal form

θ = 1.04719 \dots + πn, θ = 2.09439 \dots + πn

tan (\frac{x}{7}) + 3 = 0

3 (tan (θ) - 2) = 2 tan (θ) - 7

3 tan (θ) + 1 = 0, 0 \leq θ \leq 2 π

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

y, sin (x y) = x

What is the general solution for 3cot^2(θ)-1=0 ?
The general solution for 3cot^2(θ)-1=0 is θ=1.04719…+pin,θ=2.09439…+pin