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

The general solution for 3tan^2(θ)+1= 2/(tan^2(θ)) is θ=0.68471…+pin,θ=-0.68471…+pin

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Popular Trigonometry >

3tan^2(θ)+1= 2/(tan^2(θ))

Solution

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

Solution

θ = 0.68471 \dots + πn, θ = - 0.68471 \dots + πn

Degrees

θ = 39.23152 \dots^{\circ} + 18 0^{\circ} n, θ = - 39.23152 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (θ) = u

3 u^{2} + 1 = \frac{2}{u ^{2}}

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

3 u^{2} + 1 = \frac{2}{u ^{2}}

Multiply both sides by

u^{2}

3 u^{2} + 1 = \frac{2}{u ^{2}}

Multiply both sides by

u^{2}

3 u^{2} u^{2} + 1 \cdot u^{2} = \frac{2}{u ^{2}} u^{2}

Simplify

3 u^{2} u^{2} : 3 u^{4}

3 u^{2} u^{2} + 1 \cdot u^{2} = \frac{2}{u ^{2}} u^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u^{2} = u^{2 + 2}

= 3 u^{2 + 2}

Add the numbers:

2 + 2 = 4

= 3 u^{4}

3 u^{4} + u^{2} = 2

3 u^{4} + u^{2} = 2

Solve

3 u^{4} + u^{2} = 2 : u = \frac{2}{3}, u = - \frac{2}{3}, u = i, u = - i

3 u^{4} + u^{2} = 2

Move

2

to the left side

3 u^{4} + u^{2} = 2

Subtract

2

from both sides

3 u^{4} + u^{2} - 2 = 2 - 2

Simplify

3 u^{4} + u^{2} - 2 = 0

3 u^{4} + u^{2} - 2 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

3 v^{2} + v - 2 = 0

Solve

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

3 v^{2} + v - 2 = 0

Solve with the quadratic formula

3 v^{2} + v - 2 = 0

Quadratic Equation Formula:

For

a = 3, b = 1, c = - 2

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 2) = 5

1^{2} - 4 \cdot 3 (- 2)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 3 (- 2)

Apply rule

- (- a) = a

= 1 + 4 \cdot 3 \cdot 2

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

v_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 3}

Separate the solutions

v_{1} = \frac{- 1 + 5}{2 \cdot 3}, v_{2} = \frac{- 1 - 5}{2 \cdot 3}

v = \frac{- 1 + 5}{2 \cdot 3} : \frac{2}{3}

\frac{- 1 + 5}{2 \cdot 3}

Add/Subtract the numbers:

- 1 + 5 = 4

= \frac{4}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{4}{6}

Cancel the common factor:

2

= \frac{2}{3}

v = \frac{- 1 - 5}{2 \cdot 3} : - 1

\frac{- 1 - 5}{2 \cdot 3}

Subtract the numbers:

- 1 - 5 = - 6

= \frac{- 6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 6}{6}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

v = \frac{2}{3}, v = - 1

v = \frac{2}{3}, v = - 1

Substitute back

v = u^{2},

solve for

u

Solve

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

u^{2} = \frac{2}{3}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{2}{3}, u = - \frac{2}{3}

Solve

u^{2} = - 1 : u = i, u = - i

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

The solutions are

u = \frac{2}{3}, u = - \frac{2}{3}, u = i, u = - i

u = \frac{2}{3}, u = - \frac{2}{3}, u = i, u = - i

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

\frac{2}{u ^{2}}

and compare to zero

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{2}{3}, u = - \frac{2}{3}, u = i, u = - i

Substitute back

u = tan (θ)

tan (θ) = \frac{2}{3}, tan (θ) = - \frac{2}{3}, tan (θ) = i, tan (θ) = - i

tan (θ) = \frac{2}{3}, tan (θ) = - \frac{2}{3}, tan (θ) = i, tan (θ) = - i

tan (θ) = \frac{2}{3} : θ = arctan (\frac{2}{3}) + πn

tan (θ) = \frac{2}{3}

Apply trig inverse properties

tan (θ) = \frac{2}{3}

General solutions for

tan (θ) = \frac{2}{3}

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

θ = arctan (\frac{2}{3}) + πn

θ = arctan (\frac{2}{3}) + πn

tan (θ) = - \frac{2}{3} : θ = arctan (- \frac{2}{3}) + πn

tan (θ) = - \frac{2}{3}

Apply trig inverse properties

tan (θ) = - \frac{2}{3}

General solutions for

tan (θ) = - \frac{2}{3}

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

θ = arctan (- \frac{2}{3}) + πn

θ = arctan (- \frac{2}{3}) + πn

tan (θ) = i :

No Solution

tan (θ) = i

No Solution

tan (θ) = - i :

No Solution

tan (θ) = - i

No Solution

Combine all the solutions

θ = arctan (\frac{2}{3}) + πn, θ = arctan (- \frac{2}{3}) + πn

Show solutions in decimal form

θ = 0.68471 \dots + πn, θ = - 0.68471 \dots + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3tan^2(θ)+1= 2/(tan^2(θ)) ?
The general solution for 3tan^2(θ)+1= 2/(tan^2(θ)) is θ=0.68471…+pin,θ=-0.68471…+pin