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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan^2(x)=-1+2tan(-x)

Solution

tan^{2} (x) = - 1 + 2 tan (- x)

Solution

x = \frac{3 π}{4} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) = - 1 + 2 tan (- x)

Rewrite using trig identities

tan^{2} (x) = - 1 + 2 tan (- x)

tan^{2} (x) = - 1 + 2 (- tan (x))

Simplify

- 1 + 2 (- tan (x)) : - 1 - 2 tan (x)

- 1 + 2 (- tan (x))

Remove parentheses:

(- a) = - a

= - 1 - 2 tan (x)

tan^{2} (x) = - 1 - 2 tan (x)

tan^{2} (x) = - 1 - 2 tan (x)

Subtract

- 1 - 2 tan (x)

from both sides

tan^{2} (x) + 1 + 2 tan (x) = 0

Solve by substitution

tan^{2} (x) + 1 + 2 tan (x) = 0

Let:

tan (x) = u

u^{2} + 1 + 2 u = 0

u^{2} + 1 + 2 u = 0 : u = - 1

u^{2} + 1 + 2 u = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} + 2 u + 1 = 0

Solve with the quadratic formula

u^{2} + 2 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 2, c = 1

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot 1 = 0

2^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

u_{1, 2} = \frac{- 2 \pm 0}{2 \cdot 1}

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

\frac{- 2}{2 \cdot 1} = - 1

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

u = - 1

The solution to the quadratic equation is:

u = - 1

Substitute back

u = tan (x)

tan (x) = - 1

tan (x) = - 1

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

tan (x) = - 1

General solutions for

tan (x) = - 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}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{3 π}{4} + πn

Popular Examples

tan(2θ)-tan(θ)=0

tan (2 θ) - tan (θ) = 0

cos(t)=-1/2

cos (t) = - \frac{1}{2}

cos(2θ)+cos(θ)=0

cos (2 θ) + cos (θ) = 0

4sin(x)=2

4 sin (x) = 2

tan(4x)=1

tan (4 x) = 1

Frequently Asked Questions (FAQ)

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