What is the general solution for 1+tan^2(x)cos(x)=sec(x) ?

The general solution for 1+tan^2(x)cos(x)=sec(x) is x=2pin

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Popular >

1+tan^2(x)cos(x)=sec(x)

Solution

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

Solution

x = 2 πn

Degrees

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

Solution steps

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

Subtract

sec (x)

from both sides

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

Express with sin, cos

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

Simplify

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

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

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

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

Combine the fractions

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

Apply rule

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

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

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

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

Subtract

sin^{2} (x)

from both sides

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

Square both sides

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

Subtract

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

from both sides

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

Factor

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

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

Apply exponent rule:

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

Refine

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

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

Solving each part separately

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

cos (x) - 1 + sin^{2} (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

u - u^{2} = 0

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

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 0

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 1 + 1 = 0

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

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

= - \frac{0}{2}

Apply rule

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

= - 0

= 0

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

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 1 - 1 = - 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 = 0, u = 1

Substitute back

u = cos (x)

cos (x) = 0, cos (x) = 1

cos (x) = 0, cos (x) = 1

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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

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

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

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 1

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

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

Combine all the solutions

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

Simplify

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

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

Group like terms

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

Subtract the numbers:

- 1 - 1 = - 2

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

- 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 \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

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 2)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 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} : 1

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

Add/Subtract the numbers:

- 1 + 3 = 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 - 3}{2 \cdot 1} : - 2

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

Subtract 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

The solutions to the quadratic equation are:

u = 1, u = - 2

Substitute back

u = cos (x)

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

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

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 1

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

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

cos (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

plug in

x = \frac{π}{2} + 2 π 1

1 + tan^{2} (\frac{π}{2} + 2 π 1) cos (\frac{π}{2} + 2 π 1) = sec (\frac{π}{2} + 2 π 1)

Undefined

\Rightarrow False

Check the solution

\frac{3 π}{2} + 2 πn :

False

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

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

plug in

x = \frac{3 π}{2} + 2 π 1

1 + tan^{2} (\frac{3 π}{2} + 2 π 1) cos (\frac{3 π}{2} + 2 π 1) = sec (\frac{3 π}{2} + 2 π 1)

Undefined

\Rightarrow False

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

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

plug in

x = 2 π 1

1 + tan^{2} (2 π 1) cos (2 π 1) = sec (2 π 1)

Refine

1 = 1

\Rightarrow True

x = 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 1+tan^2(x)cos(x)=sec(x) ?
The general solution for 1+tan^2(x)cos(x)=sec(x) is x=2pin