What is the general solution for tan^2(x)+tan(x)-30=0 ?

The general solution for tan^2(x)+tan(x)-30=0 is x=1.37340…+pin,x=-1.40564…+pin

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tan^2(x)+tan(x)-30=0

Solution

tan^{2} (x) + tan (x) - 30 = 0

Solution

x = 1.37340 \dots + πn, x = - 1.40564 \dots + πn

Degrees

x = 78.69006 \dots^{\circ} + 18 0^{\circ} n, x = - 80.53767 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) + tan (x) - 30 = 0

Solve by substitution

tan^{2} (x) + tan (x) - 30 = 0

Let:

tan (x) = u

u^{2} + u - 30 = 0

u^{2} + u - 30 = 0 : u = 5, u = - 6

u^{2} + u - 30 = 0

Solve with the quadratic formula

u^{2} + u - 30 = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = - 30

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 30 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 30 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 30) = 11

1^{2} - 4 \cdot 1 \cdot (- 30)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 30)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 30

Multiply the numbers:

4 \cdot 1 \cdot 30 = 120

= 1 + 120

Add the numbers:

1 + 120 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- 1 \pm 11}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 11}{2 \cdot 1}, u_{2} = \frac{- 1 - 11}{2 \cdot 1}

u = \frac{- 1 + 11}{2 \cdot 1} : 5

\frac{- 1 + 11}{2 \cdot 1}

Add/Subtract the numbers:

- 1 + 11 = 10

= \frac{10}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{10}{2}

Divide the numbers:

\frac{10}{2} = 5

= 5

u = \frac{- 1 - 11}{2 \cdot 1} : - 6

\frac{- 1 - 11}{2 \cdot 1}

Subtract the numbers:

- 1 - 11 = - 12

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 12}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{12}{2}

Divide the numbers:

\frac{12}{2} = 6

= - 6

The solutions to the quadratic equation are:

u = 5, u = - 6

Substitute back

u = tan (x)

tan (x) = 5, tan (x) = - 6

tan (x) = 5, tan (x) = - 6

tan (x) = 5 : x = arctan (5) + πn

tan (x) = 5

Apply trig inverse properties

tan (x) = 5

General solutions for

tan (x) = 5

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

x = arctan (5) + πn

x = arctan (5) + πn

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

Combine all the solutions

x = arctan (5) + πn, x = arctan (- 6) + πn

Show solutions in decimal form

x = 1.37340 \dots + πn, x = - 1.40564 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)+tan(x)-30=0 ?
The general solution for tan^2(x)+tan(x)-30=0 is x=1.37340…+pin,x=-1.40564…+pin