What is the general solution for sec(x)-2cos(x)=1 ?

The general solution for sec(x)-2cos(x)=1 is x= pi/3+2pin,x=(5pi)/3+2pin,x=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sec(x)-2cos(x)=1

Solution

sec (x) - 2 cos (x) = 1

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = π + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sec (x) - 2 cos (x) = 1

Subtract

1

from both sides

sec (x) - 2 cos (x) - 1 = 0

Rewrite using trig identities

- 1 + sec (x) - 2 cos (x)

Use the basic trigonometric identity:

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

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

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

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

Solve by substitution

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

Let:

sec (x) = u

- 1 - \frac{2}{u} + u = 0

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

- 1 - \frac{2}{u} + u = 0

Multiply both sides by

u

- 1 - \frac{2}{u} + u = 0

Multiply both sides by

u

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

Simplify

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

Simplify

- 1 \cdot u : - u

- 1 \cdot u

Multiply:

1 \cdot u = u

= - 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

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

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

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

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

Solve

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

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

Write in the standard form

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

u^{2} - u - 2 = 0

Solve with the quadratic formula

u^{2} - u - 2 = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 2 = 8

4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 3 = 4

= \frac{4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 3 = - 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 = 2, u = - 1

u = 2, u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

- 1 - \frac{2}{u} + u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 2, u = - 1

Substitute back

u = sec (x)

sec (x) = 2, sec (x) = - 1

sec (x) = 2, sec (x) = - 1

sec (x) = 2 : x = \frac{π}{3} + 2 πn, x = \frac{5 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{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

sec (x) = - 1 : x = π + 2 πn

sec (x) = - 1

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 = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = π + 2 πn

Popular Examples

2sin(x)+3=0

2 sin (x) + 3 = 0

4cos(2θ)-10cos(θ)+14=7

4 cos (2 θ) - 10 cos (θ) + 14 = 7

2cos^2(2x)=cos(2x)

2 cos^{2} (2 x) = cos (2 x)

0=8cos(x)-8sin(x)

0 = 8 cos (x) - 8 sin (x)

sec(8x)=2

sec (8 x) = 2

Frequently Asked Questions (FAQ)

What is the general solution for sec(x)-2cos(x)=1 ?
The general solution for sec(x)-2cos(x)=1 is x= pi/3+2pin,x=(5pi)/3+2pin,x=pi+2pin