What is the general solution for -1-tan(x)=sec(x),0<= x<2pi ?

Q: What is the general solution for -1-tan(x)=sec(x),0<= x<2pi ?

The general solution for -1-tan(x)=sec(x),0<= x<2pi is x=pi

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-1-tan(x)=sec(x),0<= x<2pi

Solution

- 1 - tan (x) = sec (x), 0 \leq x < 2 π

Solution

x = π

Degrees

x = 18 0^{\circ}

Solution steps

- 1 - tan (x) = sec (x), 0 \leq x < 2 π

Subtract

sec (x)

from both sides

- 1 - tan (x) - sec (x) = 0

Express with sin, cos

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

Simplify

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

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

Combine the fractions

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

Apply rule

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

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

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

- cos (x) - sin (x) - 1 = 0

Add

sin (x)

to both sides

- cos (x) - 1 = sin (x)

Square both sides

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

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = - 1, b = cos (x)

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

Simplify

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

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

Apply rule

- (- a) = a

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

(- 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

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 \cdot cos (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos (x)

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

1 - 1 = 0

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

2 u + 2 u^{2} = 0

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

2 u + 2 u^{2} = 0

Write in the standard form

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

2 u^{2} + 2 u = 0

Solve with the quadratic formula

2 u^{2} + 2 u = 0

Quadratic Equation Formula:

For

a = 2, b = 2, c = 0

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

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

2^{2} - 4 \cdot 2 \cdot 0 = 2

2^{2} - 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

u = \frac{- 2 + 2}{2 \cdot 2} : 0

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

Add/Subtract the numbers:

- 2 + 2 = 0

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

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

= 0

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

\frac{- 2 - 2}{2 \cdot 2}

Subtract the numbers:

- 2 - 2 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

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, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x < 2 π

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

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = - 1, 0 \leq x < 2 π : x = π

cos (x) = - 1, 0 \leq x < 2 π

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 = π + 2 πn

x = π + 2 πn

Solutions for the range

0 \leq x < 2 π

x = π

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

- 1 - tan (x) = sec (x)

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

Check the solution

\frac{π}{2} :

False

\frac{π}{2}

Plug in

n = 1

\frac{π}{2}

For

- 1 - tan (x) = sec (x)

plug in

x = \frac{π}{2}

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

Refine

- \infty = \infty

\Rightarrow False

Check the solution

\frac{3 π}{2} :

True

\frac{3 π}{2}

Plug in

n = 1

\frac{3 π}{2}

For

- 1 - tan (x) = sec (x)

plug in

x = \frac{3 π}{2}

- 1 - tan (\frac{3 π}{2}) = sec (\frac{3 π}{2})

Refine

- \infty = - \infty

\Rightarrow True

Check the solution

π :

True

π

Plug in

n = 1

π

For

- 1 - tan (x) = sec (x)

plug in

x = π

- 1 - tan (π) = sec (π)

Refine

- 1 = - 1

\Rightarrow True

x = \frac{3 π}{2}, x = π

Since the equation is undefined for:

\frac{3 π}{2}

x = π

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Frequently Asked Questions (FAQ)

What is the general solution for -1-tan(x)=sec(x),0<= x<2pi ?
The general solution for -1-tan(x)=sec(x),0<= x<2pi is x=pi