What is the general solution for 2cos(x)-5=3sec(x) ?

The general solution for 2cos(x)-5=3sec(x) is x=(2pi)/3+2pin,x=(4pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos(x)-5=3sec(x)

Solution

2 cos (x) - 5 = 3 sec (x)

Solution

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (x) - 5 = 3 sec (x)

Subtract

3 sec (x)

from both sides

2 cos (x) - 5 - 3 sec (x) = 0

Rewrite using trig identities

- 5 + 2 cos (x) - 3 sec (x)

Use the basic trigonometric identity:

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

= - 5 + 2 \cdot \frac{1}{sec ( x )} - 3 sec (x)

2 \cdot \frac{1}{sec ( x )} = \frac{2}{sec ( x )}

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

Multiply fractions:

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

= \frac{1 \cdot 2}{sec ( x )}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{sec ( x )}

= - 5 + \frac{2}{sec ( x )} - 3 sec (x)

- 5 + \frac{2}{sec ( x )} - 3 sec (x) = 0

Solve by substitution

- 5 + \frac{2}{sec ( x )} - 3 sec (x) = 0

Let:

sec (x) = u

- 5 + \frac{2}{u} - 3 u = 0

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

- 5 + \frac{2}{u} - 3 u = 0

Multiply both sides by

u

- 5 + \frac{2}{u} - 3 u = 0

Multiply both sides by

u

- 5 u + \frac{2}{u} u - 3 uu = 0 \cdot u

Simplify

- 5 u + \frac{2}{u} u - 3 uu = 0 \cdot u

Simplify

\frac{2}{u} u : 2

\frac{2}{u} u

Multiply fractions:

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

= \frac{2 u}{u}

Cancel the common factor:

u

= 2

Simplify

- 3 uu : - 3 u^{2}

- 3 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= - 3 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= - 3 u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

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

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

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

Solve

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 3, b = - 5, c = 2

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

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

(- 5)^{2} - 4 (- 3) \cdot 2 = 7

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 5^{2} + 4 \cdot 3 \cdot 2

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 5 ) \pm 7}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 7}{2 ( - 3 )}, u_{2} = \frac{- ( - 5 ) - 7}{2 ( - 3 )}

u = \frac{- ( - 5 ) + 7}{2 ( - 3 )} : - 2

\frac{- ( - 5 ) + 7}{2 ( - 3 )}

Remove parentheses:

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

= \frac{5 + 7}{- 2 \cdot 3}

Add the numbers:

5 + 7 = 12

= \frac{12}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{12}{- 6}

Apply the fraction rule:

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

= - \frac{12}{6}

Divide the numbers:

\frac{12}{6} = 2

= - 2

u = \frac{- ( - 5 ) - 7}{2 ( - 3 )} : \frac{1}{3}

\frac{- ( - 5 ) - 7}{2 ( - 3 )}

Remove parentheses:

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

= \frac{5 - 7}{- 2 \cdot 3}

Subtract the numbers:

5 - 7 = - 2

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2}{- 6}

Apply the fraction rule:

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

= \frac{2}{6}

Cancel the common factor:

2

= \frac{1}{3}

The solutions to the quadratic equation are:

u = - 2, u = \frac{1}{3}

u = - 2, u = \frac{1}{3}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

- 5 + \frac{2}{u} - 3 u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 2, u = \frac{1}{3}

Substitute back

u = sec (x)

sec (x) = - 2, sec (x) = \frac{1}{3}

sec (x) = - 2, sec (x) = \frac{1}{3}

sec (x) = - 2 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

sec (x) = - 2

General solutions for

sec (x) = - 2

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 = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

sec (x) = \frac{1}{3} :

No Solution

sec (x) = \frac{1}{3}

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

No Solution

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Popular Examples

sin(3θ+pi/(18))=1,0<= θ<2pi

sin (3 θ + \frac{π}{18}) = 1, 0 \leq θ < 2 π

1=tan(c)

1 = tan (c)

sin(a)=(sqrt(3))/2

sin (a) = \frac{3}{2}

tan^2(x)=4

tan^{2} (x) = 4

sec(x)=sqrt(3)

sec (x) = 3

Frequently Asked Questions (FAQ)

What is the general solution for 2cos(x)-5=3sec(x) ?
The general solution for 2cos(x)-5=3sec(x) is x=(2pi)/3+2pin,x=(4pi)/3+2pin