What is the general solution for -2tan(θ)-4-tan^2(θ)=-3 ?

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

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-2tan(θ)-4-tan^2(θ)=-3

Solution

- 2 tan (θ) - 4 - tan^{2} (θ) = - 3

Solution

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

- 2 tan (θ) - 4 - tan^{2} (θ) = - 3

Solve by substitution

- 2 tan (θ) - 4 - tan^{2} (θ) = - 3

Let:

tan (θ) = u

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

- 2 u - 4 - u^{2} = - 3 : u = - 1

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

Move

3

to the left side

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

Add

3

to both sides

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

Simplify

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

= 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, - (- 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{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{3 π}{4} + πn

Popular Examples

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sin(x)= 1/(sec(x))

sin (x) = \frac{1}{sec ( x )}

solvefor x,1-2cos^2(x)-2cos(x)=0 solve for

x, 1 - 2 cos^{2} (x) - 2 cos (x) = 0

1=tan(x/3)

1 = tan (\frac{x}{3})

cos(12x)=0

cos (12 x) = 0

Frequently Asked Questions (FAQ)

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