What is the general solution for sec(x)-(sin(x))/(cos(x))=cos(x) ?

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

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Popular Trigonometry >

sec(x)-(sin(x))/(cos(x))=cos(x)

Solution

sec (x) - \frac{sin ( x )}{cos ( x )} = cos (x)

Solution

x = 2 πn, x = π + 2 πn

Degrees

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

Solution steps

sec (x) - \frac{sin ( x )}{cos ( x )} = cos (x)

Subtract

cos (x)

from both sides

sec (x) - \frac{sin ( x )}{cos ( x )} - cos (x) = 0

Simplify

sec (x) - \frac{sin ( x )}{cos ( x )} - cos (x) : \frac{sec ( x ) cos ( x ) - sin ( x ) - cos ^{2} ( x )}{cos ( x )}

sec (x) - \frac{sin ( x )}{cos ( x )} - cos (x)

Convert element to fraction:

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

= \frac{sec ( x ) cos ( x )}{cos ( x )} - \frac{sin ( x )}{cos ( x )} - \frac{cos ( x ) cos ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{sec ( x ) cos ( x ) - sin ( x ) - cos ( x ) cos ( x )}{cos ( x )}

sec (x) cos (x) - sin (x) - cos (x) cos (x) = sec (x) cos (x) - sin (x) - cos^{2} (x)

sec (x) cos (x) - sin (x) - cos (x) cos (x)

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

cos (x) cos (x)

Apply exponent rule:

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

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

= sec (x) cos (x) - sin (x) - cos^{2} (x)

= \frac{sec ( x ) cos ( x ) - sin ( x ) - cos ^{2} ( x )}{cos ( x )}

\frac{sec ( x ) cos ( x ) - sin ( x ) - cos ^{2} ( x )}{cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sec (x) cos (x) - sin (x) - cos^{2} (x) = 0

Express with sin, cos

\frac{1}{cos ( x )} cos (x) - sin (x) - cos^{2} (x) = 0

Simplify

\frac{1}{cos ( x )} cos (x) - sin (x) - cos^{2} (x) : 1 - sin (x) - cos^{2} (x)

\frac{1}{cos ( x )} cos (x) - sin (x) - cos^{2} (x)

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= 1

= 1 - sin (x) - cos^{2} (x)

1 - sin (x) - cos^{2} (x) = 0

Add

cos^{2} (x)

to both sides

1 - sin (x) = cos^{2} (x)

Square both sides

(1 - sin (x))^{2} = (cos^{2} (x))^{2}

Subtract

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

from both sides

(1 - sin (x))^{2} - cos^{4} (x) = 0

Factor

(1 - sin (x))^{2} - cos^{4} (x) : (1 - sin (x) + cos^{2} (x)) (1 - sin (x) - cos^{2} (x))

(1 - sin (x))^{2} - cos^{4} (x)

Apply exponent rule:

a^{b c} = (a^{b})^{c}

cos^{4} (x) = (cos^{2} (x))^{2}

= (1 - sin (x))^{2} - (cos^{2} (x))^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(1 - sin (x))^{2} - (cos^{2} (x))^{2} = ((1 - sin (x)) + cos^{2} (x)) ((1 - sin (x)) - cos^{2} (x))

= ((1 - sin (x)) + cos^{2} (x)) ((1 - sin (x)) - cos^{2} (x))

Refine

= (cos^{2} (x) - sin (x) + 1) (- cos^{2} (x) - sin (x) + 1)

(1 - sin (x) + cos^{2} (x)) (1 - sin (x) - cos^{2} (x)) = 0

Solving each part separately

1 - sin (x) + cos^{2} (x) = 0 or 1 - sin (x) - cos^{2} (x) = 0

1 - sin (x) + cos^{2} (x) = 0 : x = \frac{π}{2} + 2 πn

1 - sin (x) + cos^{2} (x) = 0

Rewrite using trig identities

1 + cos^{2} (x) - sin (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= 1 + 1 - sin^{2} (x) - sin (x)

Simplify

= - sin^{2} (x) - sin (x) + 2

2 - sin (x) - sin^{2} (x) = 0

Solve by substitution

2 - sin (x) - sin^{2} (x) = 0

Let:

sin (x) = u

2 - u - u^{2} = 0

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

2 - u - 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 ( - 1 ) \cdot 2}{2 ( - 1 )}

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

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

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

Separate the solutions

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

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

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

Remove parentheses:

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

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

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

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

Remove parentheses:

(- a) = - a, - (- 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

Substitute back

u = sin (x)

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

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

sin (x) = - 2 :

No Solution

sin (x) = - 2

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn

1 - sin (x) - cos^{2} (x) = 0 : x = \frac{π}{2} + 2 πn, x = 2 πn, x = π + 2 πn

1 - sin (x) - cos^{2} (x) = 0

Rewrite using trig identities

1 - cos^{2} (x) - sin (x)

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - cos^{2} (x) = sin^{2} (x)

= - sin (x) + sin^{2} (x)

- sin (x) + sin^{2} (x) = 0

Solve by substitution

- sin (x) + sin^{2} (x) = 0

Let:

sin (x) = u

- u + u^{2} = 0

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

- u + u^{2} = 0

Write in the standard form

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

u^{2} - u = 0

Solve with the quadratic formula

u^{2} - u = 0

Quadratic Equation Formula:

For

a = 1, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

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

(- 1)^{2} - 4 \cdot 1 \cdot 0

(- 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 0 = 0

4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 1 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 1 ) - 1}{2 \cdot 1} : 0

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 1 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 1, u = 0

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = 0

sin (x) = 1, sin (x) = 0

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, x = π + 2 πn

Combine all the solutions

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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

sec (x) - \frac{sin ( x )}{cos ( x )} = cos (x)

Remove the ones that don't agree with the equation.

Check the solution

\frac{π}{2} + 2 πn :

False

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

sec (x) - \frac{sin ( x )}{cos ( x )} = cos (x)

plug in

x = \frac{π}{2} + 2 π 1

sec (\frac{π}{2} + 2 π 1) - \frac{sin ( \frac{π}{2} + 2 π 1 )}{cos ( \frac{π}{2} + 2 π 1 )} = cos (\frac{π}{2} + 2 π 1)

Undefined

\Rightarrow False

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

sec (x) - \frac{sin ( x )}{cos ( x )} = cos (x)

plug in

x = 2 π 1

sec (2 π 1) - \frac{sin ( 2 π 1 )}{cos ( 2 π 1 )} = cos (2 π 1)

Refine

1 = 1

\Rightarrow True

Check the solution

π + 2 πn :

True

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

sec (x) - \frac{sin ( x )}{cos ( x )} = cos (x)

plug in

x = π + 2 π 1

sec (π + 2 π 1) - \frac{sin ( π + 2 π 1 )}{cos ( π + 2 π 1 )} = cos (π + 2 π 1)

Refine

- 1 = - 1

\Rightarrow True

x = 2 πn, x = π + 2 πn

Graph

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Frequently Asked Questions (FAQ)

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