What is the general solution for 1+tan^2(θ)=4tan^2(θ) ?

The general solution for 1+tan^2(θ)=4tan^2(θ) is θ=0.52359…+pin,θ=-0.52359…+pin

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Popular >

1+tan^2(θ)=4tan^2(θ)

Solution

1 + tan^{2} (θ) = 4 tan^{2} (θ)

Solution

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

1 + tan^{2} (θ) = 4 tan^{2} (θ)

Solve by substitution

1 + tan^{2} (θ) = 4 tan^{2} (θ)

Let:

tan (θ) = u

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

1 + u^{2} = 4 u^{2} : u = \frac{1}{3}, u = - \frac{1}{3}

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

Move

1

to the right side

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

Subtract

1

from both sides

1 + u^{2} - 1 = 4 u^{2} - 1

Simplify

u^{2} = 4 u^{2} - 1

u^{2} = 4 u^{2} - 1

Move

4 u^{2}

to the left side

u^{2} = 4 u^{2} - 1

Subtract

4 u^{2}

from both sides

u^{2} - 4 u^{2} = 4 u^{2} - 1 - 4 u^{2}

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

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Frequently Asked Questions (FAQ)

What is the general solution for 1+tan^2(θ)=4tan^2(θ) ?
The general solution for 1+tan^2(θ)=4tan^2(θ) is θ=0.52359…+pin,θ=-0.52359…+pin