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

The general solution for 2sec^2(x)=5tan(x) is x=1.10714…+pin,x=0.46364…+pin

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

Solution

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

Solution

x = 1.10714 \dots + πn, x = 0.46364 \dots + πn

Degrees

x = 63.43494 \dots^{\circ} + 18 0^{\circ} n, x = 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

5 tan (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Solve by substitution

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

Let:

tan (x) = u

(1 + u^{2}) \cdot 2 - 5 u = 0

(1 + u^{2}) \cdot 2 - 5 u = 0 : u = 2, u = \frac{1}{2}

(1 + u^{2}) \cdot 2 - 5 u = 0

Expand

(1 + u^{2}) \cdot 2 - 5 u : 2 + 2 u^{2} - 5 u

(1 + u^{2}) \cdot 2 - 5 u

= 2 (1 + u^{2}) - 5 u

Expand

2 (1 + u^{2}) : 2 + 2 u^{2}

2 (1 + u^{2})

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 + 2 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= 2 + 2 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

Subtract the numbers:

25 - 16 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 2}

u = \frac{- ( - 5 ) + 3}{2 \cdot 2} : 2

\frac{- ( - 5 ) + 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 + 3}{2 \cdot 2}

Add the numbers:

5 + 3 = 8

= \frac{8}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= 2

u = \frac{- ( - 5 ) - 3}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 5 ) - 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 - 3}{2 \cdot 2}

Subtract the numbers:

5 - 3 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = tan (x)

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

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

tan (x) = 2 : x = arctan (2) + πn

tan (x) = 2

Apply trig inverse properties

tan (x) = 2

General solutions for

tan (x) = 2

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

x = arctan (2) + πn

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

Show solutions in decimal form

x = 1.10714 \dots + πn, x = 0.46364 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sec^2(x)=5tan(x) ?
The general solution for 2sec^2(x)=5tan(x) is x=1.10714…+pin,x=0.46364…+pin