What is the general solution for 4tan^2(x)+15sec(x)=0 ?

The general solution for 4tan^2(x)+15sec(x)=0 is x=1.82347…+2pin,x=-1.82347…+2pin

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4tan^2(x)+15sec(x)=0

Solution

4 tan^{2} (x) + 15 sec (x) = 0

Solution

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

Degrees

x = 104.47751 \dots^{\circ} + 36 0^{\circ} n, x = - 104.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 tan^{2} (x) + 15 sec (x) = 0

Rewrite using trig identities

15 sec (x) + 4 tan^{2} (x)

Use the Pythagorean identity:

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

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

= 15 sec (x) + 4 (sec^{2} (x) - 1)

(- 1 + sec^{2} (x)) \cdot 4 + 15 sec (x) = 0

Solve by substitution

(- 1 + sec^{2} (x)) \cdot 4 + 15 sec (x) = 0

Let:

sec (x) = u

(- 1 + u^{2}) \cdot 4 + 15 u = 0

(- 1 + u^{2}) \cdot 4 + 15 u = 0 : u = \frac{1}{4}, u = - 4

(- 1 + u^{2}) \cdot 4 + 15 u = 0

Expand

(- 1 + u^{2}) \cdot 4 + 15 u : - 4 + 4 u^{2} + 15 u

(- 1 + u^{2}) \cdot 4 + 15 u

= 4 (- 1 + u^{2}) + 15 u

Expand

4 (- 1 + u^{2}) : - 4 + 4 u^{2}

4 (- 1 + u^{2})

Apply the distributive law:

a (b + c) = ab + a c

a = 4, b = - 1, c = u^{2}

= 4 (- 1) + 4 u^{2}

Apply minus-plus rules

+ (- a) = - a

= - 4 \cdot 1 + 4 u^{2}

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 u^{2}

= - 4 + 4 u^{2} + 15 u

- 4 + 4 u^{2} + 15 u = 0

Write in the standard form

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

4 u^{2} + 15 u - 4 = 0

Solve with the quadratic formula

4 u^{2} + 15 u - 4 = 0

Quadratic Equation Formula:

For

a = 4, b = 15, c = - 4

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 4 = 64

= 1 5^{2} + 64

1 5^{2} = 225

= 225 + 64

Add the numbers:

225 + 64 = 289

= 289

Factor the number:

289 = 1 7^{2}

= 1 7^{2}

Apply radical rule:

1 7^{2} = 17

= 17

u_{1, 2} = \frac{- 15 \pm 17}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 15 + 17}{2 \cdot 4}, u_{2} = \frac{- 15 - 17}{2 \cdot 4}

u = \frac{- 15 + 17}{2 \cdot 4} : \frac{1}{4}

\frac{- 15 + 17}{2 \cdot 4}

Add/Subtract the numbers:

- 15 + 17 = 2

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2}{8}

Cancel the common factor:

2

= \frac{1}{4}

u = \frac{- 15 - 17}{2 \cdot 4} : - 4

\frac{- 15 - 17}{2 \cdot 4}

Subtract the numbers:

- 15 - 17 = - 32

= \frac{- 32}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 32}{8}

Apply the fraction rule:

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

= - \frac{32}{8}

Divide the numbers:

\frac{32}{8} = 4

= - 4

The solutions to the quadratic equation are:

u = \frac{1}{4}, u = - 4

Substitute back

u = sec (x)

sec (x) = \frac{1}{4}, sec (x) = - 4

sec (x) = \frac{1}{4}, sec (x) = - 4

sec (x) = \frac{1}{4} :

No Solution

sec (x) = \frac{1}{4}

sec (x) \leq - 1 or sec (x) \geq 1

No Solution

sec (x) = - 4 : x = arcsec (- 4) + 2 πn, x = - arcsec (- 4) + 2 πn

sec (x) = - 4

Apply trig inverse properties

sec (x) = - 4

General solutions for

sec (x) = - 4

sec (x) = - a \Rightarrow x = arcsec (- a) + 2 πn, x = - arcsec (- a) + 2 πn

x = arcsec (- 4) + 2 πn, x = - arcsec (- 4) + 2 πn

x = arcsec (- 4) + 2 πn, x = - arcsec (- 4) + 2 πn

Combine all the solutions

x = arcsec (- 4) + 2 πn, x = - arcsec (- 4) + 2 πn

Show solutions in decimal form

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4tan^2(x)+15sec(x)=0 ?
The general solution for 4tan^2(x)+15sec(x)=0 is x=1.82347…+2pin,x=-1.82347…+2pin