What is the general solution for tan^2(x)-5tan(x)-9=0 ?

The general solution for tan^2(x)-5tan(x)-9=0 is x=1.41592…+pin,x=-0.95227…+pin

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tan^2(x)-5tan(x)-9=0

Solution

tan^{2} (x) - 5 tan (x) - 9 = 0

Solution

x = 1.41592 \dots + πn, x = - 0.95227 \dots + πn

Degrees

x = 81.12633 \dots^{\circ} + 18 0^{\circ} n, x = - 54.56128 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) - 5 tan (x) - 9 = 0

Solve by substitution

tan^{2} (x) - 5 tan (x) - 9 = 0

Let:

tan (x) = u

u^{2} - 5 u - 9 = 0

u^{2} - 5 u - 9 = 0 : u = \frac{5 + 61}{2}, u = \frac{5 - 61}{2}

u^{2} - 5 u - 9 = 0

Solve with the quadratic formula

u^{2} - 5 u - 9 = 0

Quadratic Equation Formula:

For

a = 1, b = - 5, c = - 9

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 9) = 61

(- 5)^{2} - 4 \cdot 1 \cdot (- 9)

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 1 \cdot 9

Apply exponent rule:

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

n

is even

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

= 5^{2} + 4 \cdot 1 \cdot 9

Multiply the numbers:

4 \cdot 1 \cdot 9 = 36

= 5^{2} + 36

5^{2} = 25

= 25 + 36

Add the numbers:

25 + 36 = 61

= 61

u_{1, 2} = \frac{- ( - 5 ) \pm 61}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 61}{2 \cdot 1}, u_{2} = \frac{- ( - 5 ) - 61}{2 \cdot 1}

u = \frac{- ( - 5 ) + 61}{2 \cdot 1} : \frac{5 + 61}{2}

\frac{- ( - 5 ) + 61}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{5 + 61}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{5 + 61}{2}

u = \frac{- ( - 5 ) - 61}{2 \cdot 1} : \frac{5 - 61}{2}

\frac{- ( - 5 ) - 61}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{5 - 61}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{5 - 61}{2}

The solutions to the quadratic equation are:

u = \frac{5 + 61}{2}, u = \frac{5 - 61}{2}

Substitute back

u = tan (x)

tan (x) = \frac{5 + 61}{2}, tan (x) = \frac{5 - 61}{2}

tan (x) = \frac{5 + 61}{2}, tan (x) = \frac{5 - 61}{2}

tan (x) = \frac{5 + 61}{2} : x = arctan (\frac{5 + 61}{2}) + πn

tan (x) = \frac{5 + 61}{2}

Apply trig inverse properties

tan (x) = \frac{5 + 61}{2}

General solutions for

tan (x) = \frac{5 + 61}{2}

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

x = arctan (\frac{5 + 61}{2}) + πn

x = arctan (\frac{5 + 61}{2}) + πn

tan (x) = \frac{5 - 61}{2} : x = arctan (\frac{5 - 61}{2}) + πn

tan (x) = \frac{5 - 61}{2}

Apply trig inverse properties

tan (x) = \frac{5 - 61}{2}

General solutions for

tan (x) = \frac{5 - 61}{2}

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

x = arctan (\frac{5 - 61}{2}) + πn

x = arctan (\frac{5 - 61}{2}) + πn

Combine all the solutions

x = arctan (\frac{5 + 61}{2}) + πn, x = arctan (\frac{5 - 61}{2}) + πn

Show solutions in decimal form

x = 1.41592 \dots + πn, x = - 0.95227 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)-5tan(x)-9=0 ?
The general solution for tan^2(x)-5tan(x)-9=0 is x=1.41592…+pin,x=-0.95227…+pin