What is the general solution for tan^2(x)-5tan(x)-6=0 ?

The general solution for tan^2(x)-5tan(x)-6=0 is x=1.40564…+pin,x=(3pi)/4+pin

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tan^2(x)-5tan(x)-6=0

Solution

tan^{2} (x) - 5 tan (x) - 6 = 0

Solution

x = 1.40564 \dots + πn, x = \frac{3 π}{4} + πn

Degrees

x = 80.53767 \dots^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) - 5 tan (x) - 6 = 0

Solve by substitution

tan^{2} (x) - 5 tan (x) - 6 = 0

Let:

tan (x) = u

u^{2} - 5 u - 6 = 0

u^{2} - 5 u - 6 = 0 : u = 6, u = - 1

u^{2} - 5 u - 6 = 0

Solve with the quadratic formula

u^{2} - 5 u - 6 = 0

Quadratic Equation Formula:

For

a = 1, b = - 5, c = - 6

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 6) = 7

(- 5)^{2} - 4 \cdot 1 \cdot (- 6)

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 1 \cdot 6

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 1 \cdot 6

Multiply the numbers:

4 \cdot 1 \cdot 6 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 5 ) \pm 7}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 7}{2 \cdot 1}, u_{2} = \frac{- ( - 5 ) - 7}{2 \cdot 1}

u = \frac{- ( - 5 ) + 7}{2 \cdot 1} : 6

\frac{- ( - 5 ) + 7}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{5 + 7}{2 \cdot 1}

Add the numbers:

5 + 7 = 12

= \frac{12}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{12}{2}

Divide the numbers:

\frac{12}{2} = 6

= 6

u = \frac{- ( - 5 ) - 7}{2 \cdot 1} : - 1

\frac{- ( - 5 ) - 7}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{5 - 7}{2 \cdot 1}

Subtract the numbers:

5 - 7 = - 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 = 6, u = - 1

Substitute back

u = tan (x)

tan (x) = 6, tan (x) = - 1

tan (x) = 6, tan (x) = - 1

tan (x) = 6 : x = arctan (6) + πn

tan (x) = 6

Apply trig inverse properties

tan (x) = 6

General solutions for

tan (x) = 6

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

x = arctan (6) + πn

x = arctan (6) + πn

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 = arctan (6) + πn, x = \frac{3 π}{4} + πn

Show solutions in decimal form

x = 1.40564 \dots + πn, x = \frac{3 π}{4} + πn

Popular Examples

tan(θ)= 12/5

tan (θ) = \frac{12}{5}

2cos^2(t)=1-cos(t)

2 cos^{2} (t) = 1 - cos (t)

sec^2(θ)+sec(θ)=0

sec^{2} (θ) + sec (θ) = 0

cos(2x)-3cos(x)+2=0

cos (2 x) - 3 cos (x) + 2 = 0

2cos(θ)+3=2

2 cos (θ) + 3 = 2

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)-5tan(x)-6=0 ?
The general solution for tan^2(x)-5tan(x)-6=0 is x=1.40564…+pin,x=(3pi)/4+pin