What is the general solution for 5sec^2(x)+tan(x)-8=0 ?

The general solution for 5sec^2(x)+tan(x)-8=0 is x=0.59787…+pin,x=-0.72223…+pin

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Popular >

5sec^2(x)+tan(x)-8=0

Solution

5 sec^{2} (x) + tan (x) - 8 = 0

Solution

x = 0.59787 \dots + πn, x = - 0.72223 \dots + πn

Degrees

x = 34.25584 \dots^{\circ} + 18 0^{\circ} n, x = - 41.38085 \dots^{\circ} + 18 0^{\circ} n

Solution steps

5 sec^{2} (x) + tan (x) - 8 = 0

Rewrite using trig identities

- 8 + tan (x) + 5 sec^{2} (x)

Use the Pythagorean identity:

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

= - 8 + tan (x) + 5 (tan^{2} (x) + 1)

Simplify

- 8 + tan (x) + 5 (tan^{2} (x) + 1) : 5 tan^{2} (x) + tan (x) - 3

- 8 + tan (x) + 5 (tan^{2} (x) + 1)

Expand

5 (tan^{2} (x) + 1) : 5 tan^{2} (x) + 5

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

Apply the distributive law:

a (b + c) = ab + a c

a = 5, b = tan^{2} (x), c = 1

= 5 tan^{2} (x) + 5 \cdot 1

Multiply the numbers:

5 \cdot 1 = 5

= 5 tan^{2} (x) + 5

= - 8 + tan (x) + 5 tan^{2} (x) + 5

Simplify

- 8 + tan (x) + 5 tan^{2} (x) + 5 : 5 tan^{2} (x) + tan (x) - 3

- 8 + tan (x) + 5 tan^{2} (x) + 5

Group like terms

= tan (x) + 5 tan^{2} (x) - 8 + 5

Add/Subtract the numbers:

- 8 + 5 = - 3

= 5 tan^{2} (x) + tan (x) - 3

= 5 tan^{2} (x) + tan (x) - 3

= 5 tan^{2} (x) + tan (x) - 3

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

Solve by substitution

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

Let:

tan (x) = u

- 3 + u + 5 u^{2} = 0

- 3 + u + 5 u^{2} = 0 : u = \frac{- 1 + 61}{10}, u = \frac{- 1 - 61}{10}

- 3 + u + 5 u^{2} = 0

Write in the standard form

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

5 u^{2} + u - 3 = 0

Solve with the quadratic formula

5 u^{2} + u - 3 = 0

Quadratic Equation Formula:

For

a = 5, b = 1, c = - 3

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

1^{2} - 4 \cdot 5 (- 3) = 61

1^{2} - 4 \cdot 5 (- 3)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 5 (- 3)

Apply rule

- (- a) = a

= 1 + 4 \cdot 5 \cdot 3

Multiply the numbers:

4 \cdot 5 \cdot 3 = 60

= 1 + 60

Add the numbers:

1 + 60 = 61

= 61

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

Separate the solutions

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

u = \frac{- 1 + 61}{2 \cdot 5} : \frac{- 1 + 61}{10}

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 1 + 61}{10}

u = \frac{- 1 - 61}{2 \cdot 5} : \frac{- 1 - 61}{10}

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 1 - 61}{10}

The solutions to the quadratic equation are:

u = \frac{- 1 + 61}{10}, u = \frac{- 1 - 61}{10}

Substitute back

u = tan (x)

tan (x) = \frac{- 1 + 61}{10}, tan (x) = \frac{- 1 - 61}{10}

tan (x) = \frac{- 1 + 61}{10}, tan (x) = \frac{- 1 - 61}{10}

tan (x) = \frac{- 1 + 61}{10} : x = arctan (\frac{- 1 + 61}{10}) + πn

tan (x) = \frac{- 1 + 61}{10}

Apply trig inverse properties

tan (x) = \frac{- 1 + 61}{10}

General solutions for

tan (x) = \frac{- 1 + 61}{10}

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

x = arctan (\frac{- 1 + 61}{10}) + πn

x = arctan (\frac{- 1 + 61}{10}) + πn

tan (x) = \frac{- 1 - 61}{10} : x = arctan (\frac{- 1 - 61}{10}) + πn

tan (x) = \frac{- 1 - 61}{10}

Apply trig inverse properties

tan (x) = \frac{- 1 - 61}{10}

General solutions for

tan (x) = \frac{- 1 - 61}{10}

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

x = arctan (\frac{- 1 - 61}{10}) + πn

x = arctan (\frac{- 1 - 61}{10}) + πn

Combine all the solutions

x = arctan (\frac{- 1 + 61}{10}) + πn, x = arctan (\frac{- 1 - 61}{10}) + πn

Show solutions in decimal form

x = 0.59787 \dots + πn, x = - 0.72223 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5sec^2(x)+tan(x)-8=0 ?
The general solution for 5sec^2(x)+tan(x)-8=0 is x=0.59787…+pin,x=-0.72223…+pin