What is the general solution for tan^2(θ)-3tan(θ)-4=0 ?

The general solution for tan^2(θ)-3tan(θ)-4=0 is θ=1.32581…+pin,θ=(3pi)/4+pin

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Popular >

tan^2(θ)-3tan(θ)-4=0

Solution

tan^{2} (θ) - 3 tan (θ) - 4 = 0

Solution

θ = 1.32581 \dots + πn, θ = \frac{3 π}{4} + πn

Degrees

θ = 75.96375 \dots^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) - 3 tan (θ) - 4 = 0

Solve by substitution

tan^{2} (θ) - 3 tan (θ) - 4 = 0

Let:

tan (θ) = u

u^{2} - 3 u - 4 = 0

u^{2} - 3 u - 4 = 0 : u = 4, u = - 1

u^{2} - 3 u - 4 = 0

Solve with the quadratic formula

u^{2} - 3 u - 4 = 0

Quadratic Equation Formula:

For

a = 1, b = - 3, c = - 4

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

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 4) = 5

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

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

Multiply the numbers:

4 \cdot 1 \cdot 4 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, u_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

u = \frac{- ( - 3 ) + 5}{2 \cdot 1} : 4

\frac{- ( - 3 ) + 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 + 5}{2 \cdot 1}

Add the numbers:

3 + 5 = 8

= \frac{8}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= 4

u = \frac{- ( - 3 ) - 5}{2 \cdot 1} : - 1

\frac{- ( - 3 ) - 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 - 5}{2 \cdot 1}

Subtract the numbers:

3 - 5 = - 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 = 4, u = - 1

Substitute back

u = tan (θ)

tan (θ) = 4, tan (θ) = - 1

tan (θ) = 4, tan (θ) = - 1

tan (θ) = 4 : θ = arctan (4) + πn

tan (θ) = 4

Apply trig inverse properties

tan (θ) = 4

General solutions for

tan (θ) = 4

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

θ = arctan (4) + πn

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

Show solutions in decimal form

θ = 1.32581 \dots + πn, θ = \frac{3 π}{4} + πn

Popular Examples

3.25=(9.8)sin(θ)

3.25 = (9.8) sin (θ)

sin(θ)=-3

sin (θ) = - 3

sin(3x-pi/3)= 1/(sqrt(2))

sin (3 x - \frac{π}{3}) = \frac{1}{2}

-cos(θ)+cos(2θ)=0

- cos (θ) + cos (2 θ) = 0

cos(x)(cos(x)+1)=0

cos (x) (cos (x) + 1) = 0

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)-3tan(θ)-4=0 ?
The general solution for tan^2(θ)-3tan(θ)-4=0 is θ=1.32581…+pin,θ=(3pi)/4+pin