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

The general solution for sec^2(x)+tan^2(x)=5tan(x) is x=1.15759…+pin,x=0.21580…+pin

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sec^2(x)+tan^2(x)=5tan(x)

Solution

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

Solution

x = 1.15759 \dots + πn, x = 0.21580 \dots + πn

Degrees

x = 66.32508 \dots^{\circ} + 18 0^{\circ} n, x = 12.36498 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

5 tan (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

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

1 + 2 tan^{2} (x) - 5 tan (x) = 0

Solve by substitution

1 + 2 tan^{2} (x) - 5 tan (x) = 0

Let:

tan (x) = u

1 + 2 u^{2} - 5 u = 0

1 + 2 u^{2} - 5 u = 0 : u = \frac{5 + 17}{4}, u = \frac{5 - 17}{4}

1 + 2 u^{2} - 5 u = 0

Write in the standard form

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

2 u^{2} - 5 u + 1 = 0

Solve with the quadratic formula

2 u^{2} - 5 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = - 5, c = 1

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

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

(- 5)^{2} - 4 \cdot 2 \cdot 1 = 17

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

Apply exponent rule:

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

n

is even

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

= 5^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 5^{2} - 8

5^{2} = 25

= 25 - 8

Subtract the numbers:

25 - 8 = 17

= 17

u_{1, 2} = \frac{- ( - 5 ) \pm 17}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 17}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 17}{2 \cdot 2}

u = \frac{- ( - 5 ) + 17}{2 \cdot 2} : \frac{5 + 17}{4}

\frac{- ( - 5 ) + 17}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 + 17}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{5 + 17}{4}

u = \frac{- ( - 5 ) - 17}{2 \cdot 2} : \frac{5 - 17}{4}

\frac{- ( - 5 ) - 17}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 - 17}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{5 - 17}{4}

The solutions to the quadratic equation are:

u = \frac{5 + 17}{4}, u = \frac{5 - 17}{4}

Substitute back

u = tan (x)

tan (x) = \frac{5 + 17}{4}, tan (x) = \frac{5 - 17}{4}

tan (x) = \frac{5 + 17}{4}, tan (x) = \frac{5 - 17}{4}

tan (x) = \frac{5 + 17}{4} : x = arctan (\frac{5 + 17}{4}) + πn

tan (x) = \frac{5 + 17}{4}

Apply trig inverse properties

tan (x) = \frac{5 + 17}{4}

General solutions for

tan (x) = \frac{5 + 17}{4}

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

x = arctan (\frac{5 + 17}{4}) + πn

x = arctan (\frac{5 + 17}{4}) + πn

tan (x) = \frac{5 - 17}{4} : x = arctan (\frac{5 - 17}{4}) + πn

tan (x) = \frac{5 - 17}{4}

Apply trig inverse properties

tan (x) = \frac{5 - 17}{4}

General solutions for

tan (x) = \frac{5 - 17}{4}

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

x = arctan (\frac{5 - 17}{4}) + πn

x = arctan (\frac{5 - 17}{4}) + πn

Combine all the solutions

x = arctan (\frac{5 + 17}{4}) + πn, x = arctan (\frac{5 - 17}{4}) + πn

Show solutions in decimal form

x = 1.15759 \dots + πn, x = 0.21580 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for sec^2(x)+tan^2(x)=5tan(x) ?
The general solution for sec^2(x)+tan^2(x)=5tan(x) is x=1.15759…+pin,x=0.21580…+pin