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

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

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-2tan(θ)=-tan^2(θ)-1

Solution

- 2 tan (θ) = - tan^{2} (θ) - 1

Solution

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

- 2 tan (θ) = - tan^{2} (θ) - 1

Solve by substitution

- 2 tan (θ) = - tan^{2} (θ) - 1

Let:

tan (θ) = u

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

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

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

Switch sides

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

Move

2 u

to the left side

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

Add

2 u

to both sides

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

Simplify

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

- 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 ( - 1 ) ( - 1 )}{2 ( - 1 )}

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

2^{2} - 4 (- 1) (- 1) = 0

2^{2} - 4 (- 1) (- 1)

Apply rule

- (- a) = a

= 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 ( - 1 )}

u = \frac{- 2}{2 ( - 1 )}

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

\frac{- 2}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \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 (θ)

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

Combine all the solutions

θ = \frac{π}{4} + πn

Graph

Popular Examples

sin^2(x)+6sin(x)=5

sin^{2} (x) + 6 sin (x) = 5

solvefor x,y+y=tan(x)solve for

x, y + y = tan (x)

cos(2x)= 1/13

cos (2 x) = \frac{1}{13}

cos(θ)=0,D=R(θ)

cos (θ) = 0, D = R (θ)

(sin(θ))/(20)=(sin(150))/(58.1826)

\frac{sin ( θ )}{20} = \frac{sin ( 15 0 ^{\circ} )}{58.1826}

Frequently Asked Questions (FAQ)

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