What is the general solution for sec(θ)-1=(sqrt(2)-1)tan(θ) ?

The general solution for sec(θ)-1=(sqrt(2)-1)tan(θ) is θ= pi/4+2pin,θ=2pin

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sec(θ)-1=(sqrt(2)-1)tan(θ)

Solution

sec (θ) - 1 = (2 - 1) tan (θ)

Solution

θ = \frac{π}{4} + 2 πn, θ = 2 πn

Degrees

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

sec (θ) - 1 = (2 - 1) tan (θ)

Subtract

(2 - 1) tan (θ)

from both sides

sec (θ) - 1 - (2 - 1) tan (θ) = 0

Express with sin, cos

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

Simplify

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

\frac{1}{cos ( θ )} - 1 - (2 - 1) \frac{sin ( θ )}{cos ( θ )}

Multiply

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

(2 - 1) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

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

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

Combine the fractions

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

Apply rule

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

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

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

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

= \frac{1 - sin ( θ ) ( 2 - 1 )}{cos ( θ )} - \frac{1 \cdot cos ( θ )}{cos ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 - sin ( θ ) ( 2 - 1 ) - 1 \cdot cos ( θ )}{cos ( θ )}

Multiply:

1 \cdot cos (θ) = cos (θ)

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

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

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

1 - (2 - 1) sin (θ) - cos (θ) = 0

Add

cos (θ)

to both sides

1 - 2 sin (θ) + sin (θ) = cos (θ)

Square both sides

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

Subtract

cos^{2} (θ)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

(1 + sin (θ) - sin (θ) 2)^{2} - (1 - sin^{2} (θ)) : 4 sin^{2} (θ) + 2 sin (θ) - 22 sin^{2} (θ) - 22 sin (θ)

(1 + sin (θ) - sin (θ) 2)^{2} - (1 - sin^{2} (θ))

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

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

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

Expand

(1 + sin (θ) - sin (θ) 2) (1 + sin (θ) - sin (θ) 2) : 2 sin (θ) - 22 sin (θ) + 3 sin^{2} (θ) - 22 sin^{2} (θ) + 1

(1 + sin (θ) - sin (θ) 2) (1 + sin (θ) - sin (θ) 2)

Distribute parentheses

= 1 \cdot 1 + 1 \cdot sin (θ) + 1 \cdot (- sin (θ) 2) + sin (θ) \cdot 1 + sin (θ) sin (θ) + sin (θ) (- sin (θ) 2) + (- sin (θ) 2) \cdot 1 + (- sin (θ) 2) sin (θ) + (- sin (θ) 2) (- sin (θ) 2)

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= 1 \cdot 1 + 1 \cdot sin (θ) - 1 \cdot 2 sin (θ) + 1 \cdot sin (θ) + sin (θ) sin (θ) - 2 sin (θ) sin (θ) - 1 \cdot 2 sin (θ) - 2 sin (θ) sin (θ) + 22 sin (θ) sin (θ)

Simplify

1 \cdot 1 + 1 \cdot sin (θ) - 1 \cdot 2 sin (θ) + 1 \cdot sin (θ) + sin (θ) sin (θ) - 2 sin (θ) sin (θ) - 1 \cdot 2 sin (θ) - 2 sin (θ) sin (θ) + 22 sin (θ) sin (θ) : 2 sin (θ) - 22 sin (θ) + 3 sin^{2} (θ) - 22 sin^{2} (θ) + 1

1 \cdot 1 + 1 \cdot sin (θ) - 1 \cdot 2 sin (θ) + 1 \cdot sin (θ) + sin (θ) sin (θ) - 2 sin (θ) sin (θ) - 1 \cdot 2 sin (θ) - 2 sin (θ) sin (θ) + 22 sin (θ) sin (θ)

Group like terms

= 1 \cdot sin (θ) - 1 \cdot 2 sin (θ) + 1 \cdot sin (θ) + sin (θ) sin (θ) - 2 sin (θ) sin (θ) - 1 \cdot 2 sin (θ) - 2 sin (θ) sin (θ) + 22 sin (θ) sin (θ) + 1 \cdot 1

Add similar elements:

- 2 sin (θ) sin (θ) - 2 sin (θ) sin (θ) = - 22 sin (θ) sin (θ)

= 1 \cdot sin (θ) - 1 \cdot 2 sin (θ) + 1 \cdot sin (θ) + sin (θ) sin (θ) - 22 sin (θ) sin (θ) - 1 \cdot 2 sin (θ) + 22 sin (θ) sin (θ) + 1 \cdot 1

Add similar elements:

- 1 \cdot 2 sin (θ) - 1 \cdot 2 sin (θ) = - 22 sin (θ)

= 1 \cdot sin (θ) - 22 sin (θ) + 1 \cdot sin (θ) + sin (θ) sin (θ) - 22 sin (θ) sin (θ) + 22 sin (θ) sin (θ) + 1 \cdot 1

Add similar elements:

1 \cdot sin (θ) + 1 \cdot sin (θ) = 2 sin (θ)

= 2 sin (θ) - 22 sin (θ) + sin (θ) sin (θ) - 22 sin (θ) sin (θ) + 22 sin (θ) sin (θ) + 1 \cdot 1

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

22 sin (θ) sin (θ) = 22 sin^{2} (θ)

22 sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= 22 sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 22 sin^{2} (θ)

22 sin (θ) sin (θ) = 2 sin^{2} (θ)

22 sin (θ) sin (θ)

Apply radical rule:

a a = a

22 = 2

= 2 sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= 2 sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 2 sin^{2} (θ)

1 \cdot 1 = 1

1 \cdot 1

Multiply the numbers:

1 \cdot 1 = 1

= 1

= 2 sin (θ) - 22 sin (θ) + sin^{2} (θ) - 22 sin^{2} (θ) + 2 sin^{2} (θ) + 1

Add similar elements:

sin^{2} (θ) + 2 sin^{2} (θ) = 3 sin^{2} (θ)

= 2 sin (θ) - 22 sin (θ) + 3 sin^{2} (θ) - 22 sin^{2} (θ) + 1

= 2 sin (θ) - 22 sin (θ) + 3 sin^{2} (θ) - 22 sin^{2} (θ) + 1

= 2 sin (θ) - 22 sin (θ) + 3 sin^{2} (θ) - 22 sin^{2} (θ) + 1 - (1 - sin^{2} (θ))

- (1 - sin^{2} (θ)) : - 1 + sin^{2} (θ)

- (1 - sin^{2} (θ))

Distribute parentheses

= - (1) - (- sin^{2} (θ))

Apply minus-plus rules

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

= - 1 + sin^{2} (θ)

= 2 sin (θ) - 22 sin (θ) + 3 sin^{2} (θ) - 22 sin^{2} (θ) + 1 - 1 + sin^{2} (θ)

Simplify

2 sin (θ) - 22 sin (θ) + 3 sin^{2} (θ) - 22 sin^{2} (θ) + 1 - 1 + sin^{2} (θ) : 4 sin^{2} (θ) + 2 sin (θ) - 22 sin^{2} (θ) - 22 sin (θ)

2 sin (θ) - 22 sin (θ) + 3 sin^{2} (θ) - 22 sin^{2} (θ) + 1 - 1 + sin^{2} (θ)

Group like terms

= 2 sin (θ) - 22 sin (θ) + 3 sin^{2} (θ) - 22 sin^{2} (θ) + sin^{2} (θ) + 1 - 1

Add similar elements:

3 sin^{2} (θ) + sin^{2} (θ) = 4 sin^{2} (θ)

= 2 sin (θ) - 22 sin (θ) + 4 sin^{2} (θ) - 22 sin^{2} (θ) + 1 - 1

1 - 1 = 0

= 4 sin^{2} (θ) + 2 sin (θ) - 22 sin^{2} (θ) - 22 sin (θ)

= 4 sin^{2} (θ) + 2 sin (θ) - 22 sin^{2} (θ) - 22 sin (θ)

= 4 sin^{2} (θ) + 2 sin (θ) - 22 sin^{2} (θ) - 22 sin (θ)

2 sin (θ) + 4 sin^{2} (θ) - 2 sin (θ) 2 - 2 sin^{2} (θ) 2 = 0

Solve by substitution

2 sin (θ) + 4 sin^{2} (θ) - 2 sin (θ) 2 - 2 sin^{2} (θ) 2 = 0

Let:

sin (θ) = u

2 u + 4 u^{2} - 2 u 2 - 2 u^{2} 2 = 0

2 u + 4 u^{2} - 2 u 2 - 2 u^{2} 2 = 0 : u = \frac{2}{2}, u = 0

2 u + 4 u^{2} - 2 u 2 - 2 u^{2} 2 = 0

Write in the standard form

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

(4 - 22) u^{2} + (2 - 22) u = 0

Solve with the quadratic formula

(4 - 22) u^{2} + (2 - 22) u = 0

Quadratic Equation Formula:

For

a = 4 - 22, b = 2 - 22, c = 0

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

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

(2 - 22)^{2} - 4 (4 - 22) \cdot 0 = 22 - 2

(2 - 22)^{2} - 4 (4 - 22) \cdot 0

Apply rule

0 \cdot a = 0

= (2 - 22)^{2} - 0

(2 - 22)^{2} - 0 = (2 - 22)^{2}

= (2 - 22)^{2}

Expand

(2 - 22)^{2} : 12 - 82

(2 - 22)^{2}

Apply Perfect Square Formula:

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

a = 2, b = 22

= 2^{2} - 2 \cdot 2 \cdot 22 + (22)^{2}

Simplify

2^{2} - 2 \cdot 2 \cdot 22 + (22)^{2} : 12 - 82

2^{2} - 2 \cdot 2 \cdot 22 + (22)^{2}

2^{2} = 4

2^{2}

2^{2} = 4

= 4

2 \cdot 2 \cdot 22 = 82

2 \cdot 2 \cdot 22

Multiply the numbers:

2 \cdot 2 \cdot 2 = 8

= 82

(22)^{2} = 8

(22)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} (2)^{2}

(2)^{2} : 2

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

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

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 2

= 2^{2} \cdot 2

Apply exponent rule:

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

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

= 2^{2 + 1}

Add the numbers:

2 + 1 = 3

= 2^{3}

2^{3} = 8

= 8

= 4 - 82 + 8

Add the numbers:

4 + 8 = 12

= 12 - 82

= 12 - 82

= 12 - 82

= 4 - 82 + 8

= 2 \cdot 2 - 82 + 8

= (2)^{2} (2)^{2} - 82 + (8)^{2}

8 = 22

8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

= (2)^{2} (2)^{2} - 82 + (22)^{2}

222 \cdot 22 = 82

222 \cdot 22

Multiply the numbers:

2 \cdot 2 = 4

= 4222

Apply exponent rule:

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

222 = 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{\frac{1}{2} + \frac{1}{2} + \frac{1}{2}}

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

2^{\frac{1}{2} + \frac{1}{2} + \frac{1}{2}} = 22

2^{\frac{1}{2} + \frac{1}{2} + \frac{1}{2}}

Combine the fractions

\frac{1}{2} + \frac{1}{2} + \frac{1}{2} : \frac{3}{2}

Apply rule

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

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

Add the numbers:

1 + 1 + 1 = 3

= \frac{3}{2}

= 2^{\frac{3}{2}}

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

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

Apply exponent rule:

x^{a + b} = x^{a} x^{b}

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

Refine

= 22

= 4 \cdot 22

Multiply the numbers:

4 \cdot 2 = 8

= 82

= (22)^{2} - 222 \cdot 22 + (22)^{2}

Apply Perfect Square Formula:

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

(22)^{2} - 222 \cdot 22 + (22)^{2} = (22 - 22)^{2}

= (22 - 22)^{2}

Apply exponent rule:

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

n

is even

(22 - 22)^{2} = (22 - 22)^{2}

= (22 - 22)^{2}

Apply radical rule:

(22 - 22)^{2} = 22 - 22

= 22 - 22

Apply radical rule:

a a = a

22 = 2

= 22 - 2

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

Separate the solutions

u_{1} = \frac{- ( 2 - 2 2 ) + 2 2 - 2}{2 ( 4 - 2 2 )}, u_{2} = \frac{- ( 2 - 2 2 ) - ( 2 2 - 2 )}{2 ( 4 - 2 2 )}

u = \frac{- ( 2 - 2 2 ) + 2 2 - 2}{2 ( 4 - 2 2 )} : \frac{2}{2}

\frac{- ( 2 - 2 2 ) + 2 2 - 2}{2 ( 4 - 2 2 )}

Expand

- (2 - 22) + 22 - 2 : 42 - 4

- (2 - 22) + 22 - 2

- (2 - 22) : - 2 + 22

- (2 - 22)

Distribute parentheses

= - (2) - (- 22)

Apply minus-plus rules

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

= - 2 + 22

= - 2 + 22 + 22 - 2

Simplify

- 2 + 22 + 22 - 2 : 42 - 4

- 2 + 22 + 22 - 2

Add similar elements:

22 + 22 = 42

= - 2 + 42 - 2

Subtract the numbers:

- 2 - 2 = - 4

= 42 - 4

= 42 - 4

= \frac{4 2 - 4}{2 ( 4 - 2 2 )}

Factor

42 - 4 : 4 (2 - 1)

42 - 4

Rewrite as

= 42 - 4 \cdot 1

Factor out common term

4

= 4 (2 - 1)

= \frac{4 ( 2 - 1 )}{2 ( 4 - 2 2 )}

Factor

2 (4 - 22) : 4 (2 - 2)

2 (4 - 22)

Factor

4 - 22 : 2 (2 - 2)

4 - 22

Rewrite as

= 2 \cdot 2 - 22

Factor out common term

2

= 2 (2 - 2)

= 2 \cdot 2 (2 - 2)

Refine

= 4 (2 - 2)

= \frac{4 ( 2 - 1 )}{4 ( 2 - 2 )}

Divide the numbers:

\frac{4}{4} = 1

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

Remove parentheses:

(a) = a

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

Factor

2 - 2 : 2 (2 - 1)

2 - 2

2 = 22

= 22 - 2

Factor out common term

2

= 2 (2 - 1)

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

Cancel the common factor:

2 - 1

= \frac{1}{2}

Rationalize

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{2}{2}

u = \frac{- ( 2 - 2 2 ) - ( 2 2 - 2 )}{2 ( 4 - 2 2 )} : 0

\frac{- ( 2 - 2 2 ) - ( 2 2 - 2 )}{2 ( 4 - 2 2 )}

Expand

- (2 - 22) - (22 - 2) : 0

- (2 - 22) - (22 - 2)

- (2 - 22) : - 2 + 22

- (2 - 22)

Distribute parentheses

= - (2) - (- 22)

Apply minus-plus rules

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

= - 2 + 22

= - 2 + 22 - (22 - 2)

- (22 - 2) : - 22 + 2

- (22 - 2)

Distribute parentheses

= - (22) - (- 2)

Apply minus-plus rules

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

= - 22 + 2

= - 2 + 22 - 22 + 2

Simplify

- 2 + 22 - 22 + 2 : 0

- 2 + 22 - 22 + 2

Add similar elements:

22 - 22 = 0

= - 2 + 2

Add/Subtract the numbers:

- 2 + 2 = 0

= 0

= 0

= \frac{0}{2 ( 4 - 2 2 )}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = \frac{2}{2}, u = 0

Substitute back

u = sin (θ)

sin (θ) = \frac{2}{2}, sin (θ) = 0

sin (θ) = \frac{2}{2}, sin (θ) = 0

sin (θ) = \frac{2}{2} : θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

sin (θ) = \frac{2}{2}

General solutions for

sin (θ) = \frac{2}{2}

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}

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

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

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

sec (θ) - 1 = (2 - 1) tan (θ)

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

Check the solution

\frac{π}{4} + 2 πn :

True

\frac{π}{4} + 2 πn

Plug in

n = 1

\frac{π}{4} + 2 π 1

For

sec (θ) - 1 = (2 - 1) tan (θ)

plug in

θ = \frac{π}{4} + 2 π 1

sec (\frac{π}{4} + 2 π 1) - 1 = (2 - 1) tan (\frac{π}{4} + 2 π 1)

Refine

0.41421 \dots = 0.41421 \dots

\Rightarrow True

Check the solution

\frac{3 π}{4} + 2 πn :

False

\frac{3 π}{4} + 2 πn

Plug in

n = 1

\frac{3 π}{4} + 2 π 1

For

sec (θ) - 1 = (2 - 1) tan (θ)

plug in

θ = \frac{3 π}{4} + 2 π 1

sec (\frac{3 π}{4} + 2 π 1) - 1 = (2 - 1) tan (\frac{3 π}{4} + 2 π 1)

Refine

- 2.41421 \dots = - 0.41421 \dots

\Rightarrow False

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

sec (θ) - 1 = (2 - 1) tan (θ)

plug in

θ = 2 π 1

sec (2 π 1) - 1 = (2 - 1) tan (2 π 1)

Refine

0 = 0

\Rightarrow True

Check the solution

π + 2 πn :

False

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

sec (θ) - 1 = (2 - 1) tan (θ)

plug in

θ = π + 2 π 1

sec (π + 2 π 1) - 1 = (2 - 1) tan (π + 2 π 1)

Refine

- 2 = 0

\Rightarrow False

θ = \frac{π}{4} + 2 πn, θ = 2 πn

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2*cos^2(x)-cos(x)-1=0,0<= x<= pi

2 \cdot cos^{2} (x) - cos (x) - 1 = 0, 0 \leq x \leq π

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2 cos^{3} (x) - cos^{2} (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for sec(θ)-1=(sqrt(2)-1)tan(θ) ?
The general solution for sec(θ)-1=(sqrt(2)-1)tan(θ) is θ= pi/4+2pin,θ=2pin