What is the general solution for 6sec^2(x)+tan(x)-7=0 ?

The general solution for 6sec^2(x)+tan(x)-7=0 is x=0.32175…+pin,x=-0.46364…+pin

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Popular Trigonometry >

6sec^2(x)+tan(x)-7=0

Solution

6 sec^{2} (x) + tan (x) - 7 = 0

Solution

x = 0.32175 \dots + πn, x = - 0.46364 \dots + πn

Degrees

x = 18.43494 \dots^{\circ} + 18 0^{\circ} n, x = - 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

6 sec^{2} (x) + tan (x) - 7 = 0

Rewrite using trig identities

- 7 + tan (x) + 6 sec^{2} (x)

Use the Pythagorean identity:

sec^{2} (x) = tan^{2} (x) + 1

= - 7 + tan (x) + 6 (tan^{2} (x) + 1)

Simplify

- 7 + tan (x) + 6 (tan^{2} (x) + 1) : 6 tan^{2} (x) + tan (x) - 1

- 7 + tan (x) + 6 (tan^{2} (x) + 1)

Expand

6 (tan^{2} (x) + 1) : 6 tan^{2} (x) + 6

6 (tan^{2} (x) + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = 6, b = tan^{2} (x), c = 1

= 6 tan^{2} (x) + 6 \cdot 1

Multiply the numbers:

6 \cdot 1 = 6

= 6 tan^{2} (x) + 6

= - 7 + tan (x) + 6 tan^{2} (x) + 6

Simplify

- 7 + tan (x) + 6 tan^{2} (x) + 6 : 6 tan^{2} (x) + tan (x) - 1

- 7 + tan (x) + 6 tan^{2} (x) + 6

Group like terms

= tan (x) + 6 tan^{2} (x) - 7 + 6

Add/Subtract the numbers:

- 7 + 6 = - 1

= 6 tan^{2} (x) + tan (x) - 1

= 6 tan^{2} (x) + tan (x) - 1

= 6 tan^{2} (x) + tan (x) - 1

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

Solve by substitution

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

Let:

tan (x) = u

- 1 + u + 6 u^{2} = 0

- 1 + u + 6 u^{2} = 0 : u = \frac{1}{3}, u = - \frac{1}{2}

- 1 + u + 6 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

6 u^{2} + u - 1 = 0

Solve with the quadratic formula

6 u^{2} + u - 1 = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 6 (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 6 \cdot 1

Multiply the numbers:

4 \cdot 6 \cdot 1 = 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 6}, u_{2} = \frac{- 1 - 5}{2 \cdot 6}

u = \frac{- 1 + 5}{2 \cdot 6} : \frac{1}{3}

\frac{- 1 + 5}{2 \cdot 6}

Add/Subtract the numbers:

- 1 + 5 = 4

= \frac{4}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{4}{12}

Cancel the common factor:

4

= \frac{1}{3}

u = \frac{- 1 - 5}{2 \cdot 6} : - \frac{1}{2}

\frac{- 1 - 5}{2 \cdot 6}

Subtract the numbers:

- 1 - 5 = - 6

= \frac{- 6}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{12}

Apply the fraction rule:

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

= - \frac{6}{12}

Cancel the common factor:

6

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = \frac{1}{3}, u = - \frac{1}{2}

Substitute back

u = tan (x)

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{2}

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{2}

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

tan (x) = \frac{1}{3}

Apply trig inverse properties

tan (x) = \frac{1}{3}

General solutions for

tan (x) = \frac{1}{3}

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

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

tan (x) = - \frac{1}{2} : x = arctan (- \frac{1}{2}) + πn

tan (x) = - \frac{1}{2}

Apply trig inverse properties

tan (x) = - \frac{1}{2}

General solutions for

tan (x) = - \frac{1}{2}

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

x = arctan (- \frac{1}{2}) + πn

x = arctan (- \frac{1}{2}) + πn

Combine all the solutions

x = arctan (\frac{1}{3}) + πn, x = arctan (- \frac{1}{2}) + πn

Show solutions in decimal form

x = 0.32175 \dots + πn, x = - 0.46364 \dots + πn

Graph

Popular Examples

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Frequently Asked Questions (FAQ)

What is the general solution for 6sec^2(x)+tan(x)-7=0 ?
The general solution for 6sec^2(x)+tan(x)-7=0 is x=0.32175…+pin,x=-0.46364…+pin