What is the general solution for tan^2(θ)-2tan(θ)-3=0 ?

The general solution for tan^2(θ)-2tan(θ)-3=0 is θ=1.24904…+pin,θ=(3pi)/4+pin

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tan^2(θ)-2tan(θ)-3=0

Solution

tan^{2} (θ) - 2 tan (θ) - 3 = 0

Solution

θ = 1.24904 \dots + πn, θ = \frac{3 π}{4} + πn

Degrees

θ = 71.56505 \dots^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) - 2 tan (θ) - 3 = 0

Solve by substitution

tan^{2} (θ) - 2 tan (θ) - 3 = 0

Let:

tan (θ) = u

u^{2} - 2 u - 3 = 0

u^{2} - 2 u - 3 = 0 : u = 3, u = - 1

u^{2} - 2 u - 3 = 0

Solve with the quadratic formula

u^{2} - 2 u - 3 = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = - 3

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 3) = 4

(- 2)^{2} - 4 \cdot 1 \cdot (- 3)

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 1 \cdot 3

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

Add the numbers:

4 + 12 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 4}{2 \cdot 1}, u_{2} = \frac{- ( - 2 ) - 4}{2 \cdot 1}

u = \frac{- ( - 2 ) + 4}{2 \cdot 1} : 3

\frac{- ( - 2 ) + 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 + 4}{2 \cdot 1}

Add the numbers:

2 + 4 = 6

= \frac{6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

u = \frac{- ( - 2 ) - 4}{2 \cdot 1} : - 1

\frac{- ( - 2 ) - 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 - 4}{2 \cdot 1}

Subtract the numbers:

2 - 4 = - 2

= \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

The solutions to the quadratic equation are:

u = 3, u = - 1

Substitute back

u = tan (θ)

tan (θ) = 3, tan (θ) = - 1

tan (θ) = 3, tan (θ) = - 1

tan (θ) = 3 : θ = arctan (3) + πn

tan (θ) = 3

Apply trig inverse properties

tan (θ) = 3

General solutions for

tan (θ) = 3

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (3) + πn

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

Show solutions in decimal form

θ = 1.24904 \dots + πn, θ = \frac{3 π}{4} + πn

Popular Examples

cos^3(x)+sin^2(x)*cos(x)=-1/2

cos^{3} (x) + sin^{2} (x) \cdot cos (x) = - \frac{1}{2}

cos(3A)+sin(A)=0

cos (3 A) + sin (A) = 0

0=1-sec^2(x)

0 = 1 - sec^{2} (x)

sin(θ)=0.1

sin (θ) = 0.1

2/(sin(x))+10=6

\frac{2}{sin ( x )} + 10 = 6

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)-2tan(θ)-3=0 ?
The general solution for tan^2(θ)-2tan(θ)-3=0 is θ=1.24904…+pin,θ=(3pi)/4+pin