What is the general solution for tan^2(x)-7tan(x)-8=0 ?

The general solution for tan^2(x)-7tan(x)-8=0 is x=1.44644…+pin,x=(3pi)/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan^2(x)-7tan(x)-8=0

Solution

tan^{2} (x) - 7 tan (x) - 8 = 0

Solution

x = 1.44644 \dots + πn, x = \frac{3 π}{4} + πn

Degrees

x = 82.87498 \dots^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) - 7 tan (x) - 8 = 0

Solve by substitution

tan^{2} (x) - 7 tan (x) - 8 = 0

Let:

tan (x) = u

u^{2} - 7 u - 8 = 0

u^{2} - 7 u - 8 = 0 : u = 8, u = - 1

u^{2} - 7 u - 8 = 0

Solve with the quadratic formula

u^{2} - 7 u - 8 = 0

Quadratic Equation Formula:

For

a = 1, b = - 7, c = - 8

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot (- 8) = 9

(- 7)^{2} - 4 \cdot 1 \cdot (- 8)

Apply rule

- (- a) = a

= (- 7)^{2} + 4 \cdot 1 \cdot 8

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} + 4 \cdot 1 \cdot 8

Multiply the numbers:

4 \cdot 1 \cdot 8 = 32

= 7^{2} + 32

7^{2} = 49

= 49 + 32

Add the numbers:

49 + 32 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 7 ) \pm 9}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 9}{2 \cdot 1}, u_{2} = \frac{- ( - 7 ) - 9}{2 \cdot 1}

u = \frac{- ( - 7 ) + 9}{2 \cdot 1} : 8

\frac{- ( - 7 ) + 9}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{7 + 9}{2 \cdot 1}

Add the numbers:

7 + 9 = 16

= \frac{16}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{16}{2}

Divide the numbers:

\frac{16}{2} = 8

= 8

u = \frac{- ( - 7 ) - 9}{2 \cdot 1} : - 1

\frac{- ( - 7 ) - 9}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{7 - 9}{2 \cdot 1}

Subtract the numbers:

7 - 9 = - 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 = 8, u = - 1

Substitute back

u = tan (x)

tan (x) = 8, tan (x) = - 1

tan (x) = 8, tan (x) = - 1

tan (x) = 8 : x = arctan (8) + πn

tan (x) = 8

Apply trig inverse properties

tan (x) = 8

General solutions for

tan (x) = 8

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

x = arctan (8) + πn

x = arctan (8) + 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 (8) + πn, x = \frac{3 π}{4} + πn

Show solutions in decimal form

x = 1.44644 \dots + πn, x = \frac{3 π}{4} + πn

Popular Examples

5tan(x)sin(x)=-7sin(x)

5 tan (x) sin (x) = - 7 sin (x)

tan(x+pi)-cos(x+pi/2)=0

tan (x + π) - cos (x + \frac{π}{2}) = 0

2cos^2(θ)-9cos(θ)+4=0

2 cos^{2} (θ) - 9 cos (θ) + 4 = 0

cos(2x+pi/4)=(sqrt(2))/2

cos (2 x + \frac{π}{4}) = \frac{2}{2}

solvefor x,tan(x)=1 solve for

x, tan (x) = 1

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)-7tan(x)-8=0 ?
The general solution for tan^2(x)-7tan(x)-8=0 is x=1.44644…+pin,x=(3pi)/4+pin