What is the general solution for 4cos(x)=sec(x)+3,0<= x<2pi ?

Q: What is the general solution for 4cos(x)=sec(x)+3,0<= x<2pi ?

The general solution for 4cos(x)=sec(x)+3,0<= x<2pi is x=1.82347…,x=-1.82347…+2pi,x=0

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4cos(x)=sec(x)+3,0<= x<2pi

Solution

4 cos (x) = sec (x) + 3, 0 \leq x < 2 π

Solution

x = 1.82347 \dots, x = - 1.82347 \dots + 2 π, x = 0

Degrees

x = 104.47751 \dots^{\circ}, x = 255.52248 \dots^{\circ}, x = 0^{\circ}

Solution steps

4 cos (x) = sec (x) + 3, 0 \leq x < 2 π

Subtract

sec (x) + 3

from both sides

4 cos (x) - sec (x) - 3 = 0

Rewrite using trig identities

- 3 - sec (x) + 4 cos (x)

Use the basic trigonometric identity:

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

= - 3 - sec (x) + 4 \cdot \frac{1}{sec ( x )}

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Multiply the numbers:

1 \cdot 4 = 4

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

= - 3 - sec (x) + \frac{4}{sec ( x )}

- 3 + \frac{4}{sec ( x )} - sec (x) = 0

Solve by substitution

- 3 + \frac{4}{sec ( x )} - sec (x) = 0

Let:

sec (x) = u

- 3 + \frac{4}{u} - u = 0

- 3 + \frac{4}{u} - u = 0 : u = - 4, u = 1

- 3 + \frac{4}{u} - u = 0

Multiply both sides by

u

- 3 + \frac{4}{u} - u = 0

Multiply both sides by

u

- 3 u + \frac{4}{u} u - uu = 0 \cdot u

Simplify

- 3 u + \frac{4}{u} u - uu = 0 \cdot u

Simplify

\frac{4}{u} u : 4

\frac{4}{u} u

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{4 u}{u}

Cancel the common factor:

u

= 4

Simplify

- uu : - u^{2}

- uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= - u^{1 + 1}

Add the numbers:

1 + 1 = 2

= - u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

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

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

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

Solve

- 3 u + 4 - u^{2} = 0 : u = - 4, u = 1

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

Write in the standard form

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

- u^{2} - 3 u + 4 = 0

Solve with the quadratic formula

- u^{2} - 3 u + 4 = 0

Quadratic Equation Formula:

For

a = - 1, b = - 3, c = 4

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 1 \cdot 4

Multiply the numbers:

4 \cdot 1 \cdot 4 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

n a^{n} = a

5^{2} = 5

= 5

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

Separate the solutions

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

u = \frac{- ( - 3 ) + 5}{2 ( - 1 )} : - 4

\frac{- ( - 3 ) + 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a, - (- a) = a

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

Add the numbers:

3 + 5 = 8

= \frac{8}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{8}{- 2}

Apply the fraction rule:

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

= - \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= - 4

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

\frac{- ( - 3 ) - 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{3 - 5}{- 2 \cdot 1}

Subtract the numbers:

3 - 5 = - 2

= \frac{- 2}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

Apply the fraction rule:

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

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - 4, u = 1

u = - 4, u = 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

- 3 + \frac{4}{u} - u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 4, u = 1

Substitute back

u = sec (x)

sec (x) = - 4, sec (x) = 1

sec (x) = - 4, sec (x) = 1

sec (x) = - 4, 0 \leq x < 2 π : x = arcsec (- 4), x = - arcsec (- 4) + 2 π

sec (x) = - 4, 0 \leq x < 2 π

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

Solutions for the range

0 \leq x < 2 π

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

sec (x) = 1, 0 \leq x < 2 π : x = 0

sec (x) = 1, 0 \leq x < 2 π

General solutions for

sec (x) = 1

sec (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Solutions for the range

0 \leq x < 2 π

x = 0

Combine all the solutions

x = arcsec (- 4), x = - arcsec (- 4) + 2 π, x = 0

Show solutions in decimal form

x = 1.82347 \dots, x = - 1.82347 \dots + 2 π, x = 0

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos(x)=sec(x)+3,0<= x<2pi ?
The general solution for 4cos(x)=sec(x)+3,0<= x<2pi is x=1.82347…,x=-1.82347…+2pi,x=0