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

The general solution for 1-2tan^2(θ)=-tan^2(θ) is θ= pi/4+pin,θ=(3pi)/4+pin

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1-2tan^2(θ)=-tan^2(θ)

Solution

1 - 2 tan^{2} (θ) = - tan^{2} (θ)

Solution

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

1 - 2 tan^{2} (θ) = - tan^{2} (θ)

Solve by substitution

1 - 2 tan^{2} (θ) = - tan^{2} (θ)

Let:

tan (θ) = u

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

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

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

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

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

Move

u^{2}

to the left side

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

Add

u^{2}

to both sides

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

Simplify

- u^{2} = - 1

- u^{2} = - 1

Divide both sides by

- 1

- u^{2} = - 1

Divide both sides by

- 1

\frac{- u ^{2}}{- 1} = \frac{- 1}{- 1}

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

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

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = tan (θ)

tan (θ) = 1, tan (θ) = - 1

tan (θ) = 1, tan (θ) = - 1

tan (θ) = 1 : θ = \frac{π}{4} + πn

tan (θ) = 1

General solutions for

tan (θ) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

tan (θ) = - 1 : θ = \frac{3 π}{4} + πn

tan (θ) = - 1

General solutions for

tan (θ) = - 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Combine all the solutions

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

Popular Examples

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cos (a) = \frac{1}{3}

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tan(2x)+1=0

tan (2 x) + 1 = 0

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sin (x) - 2 = - sin (x)

Frequently Asked Questions (FAQ)

What is the general solution for 1-2tan^2(θ)=-tan^2(θ) ?
The general solution for 1-2tan^2(θ)=-tan^2(θ) is θ= pi/4+pin,θ=(3pi)/4+pin