What is the general solution for sec^2(x)=(sqrt(3)+1)tan(x)-sqrt(3)+1 ?

The general solution for sec^2(x)=(sqrt(3)+1)tan(x)-sqrt(3)+1 is x= pi/3+pin,x= pi/4+pin

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sec^2(x)=(sqrt(3)+1)tan(x)-sqrt(3)+1

Solution

sec^{2} (x) = (3 + 1) tan (x) - 3 + 1

Solution

x = \frac{π}{3} + πn, x = \frac{π}{4} + πn

Degrees

x = 6 0^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

sec^{2} (x) = (3 + 1) tan (x) - 3 + 1

Subtract

(3 + 1) tan (x) - 3 + 1

from both sides

sec^{2} (x) - 3 tan (x) - tan (x) + 3 - 1 = 0

Rewrite using trig identities

- 1 + sec^{2} (x) + 3 - tan (x) - 3 tan (x)

Use the Pythagorean identity:

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

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

= 3 - tan (x) - 3 tan (x) + tan^{2} (x)

3 - tan (x) + tan^{2} (x) - 3 tan (x) = 0

Solve by substitution

3 - tan (x) + tan^{2} (x) - 3 tan (x) = 0

Let:

tan (x) = u

3 - u + u^{2} - 3 u = 0

3 - u + u^{2} - 3 u = 0 : u = 3, u = 1

3 - u + u^{2} - 3 u = 0

Write in the standard form

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

u^{2} - (1 + 3) u + 3 = 0

Solve with the quadratic formula

u^{2} - (1 + 3) u + 3 = 0

Quadratic Equation Formula:

For

a = 1, b = - 1 - 3, c = 3

u_{1, 2} = \frac{- ( - 1 - 3 ) \pm ( - 1 - 3 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 1 - 3 ) \pm ( - 1 - 3 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

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

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

Multiply the numbers:

4 \cdot 1 = 4

= (- 1 - 3)^{2} - 43

Expand

(- 1 - 3)^{2} - 43 : 4 - 23

(- 1 - 3)^{2} - 43

(- 1 - 3)^{2} : 4 + 23

Apply Perfect Square Formula:

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

a = - 1, b = 3

= (- 1)^{2} - 2 (- 1) 3 + (3)^{2}

Simplify

(- 1)^{2} - 2 (- 1) 3 + (3)^{2} : 4 + 23

(- 1)^{2} - 2 (- 1) 3 + (3)^{2}

Apply rule

- (- a) = a

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

2 \cdot 1 \cdot 3 = 23

2 \cdot 1 \cdot 3

Multiply the numbers:

2 \cdot 1 = 2

= 23

(3)^{2} = 3

(3)^{2}

Apply radical rule:

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

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 3^{\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

= 3

= 1 + 23 + 3

Add the numbers:

1 + 3 = 4

= 4 + 23

= 4 + 23

= 4 + 23 - 43

Add similar elements:

23 - 43 = - 23

= 4 - 23

= 4 - 23

= 3 - 23 + 1

= (3)^{2} - 23 + (1)^{2}

1 = 1

1

Apply rule

1 = 1

= 1

= (3)^{2} - 23 + 1^{2}

23 \cdot 1 = 23

23 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 23

= (3)^{2} - 23 \cdot 1 + 1^{2}

Apply Perfect Square Formula:

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

(3)^{2} - 23 \cdot 1 + 1^{2} = (3 - 1)^{2}

= (3 - 1)^{2}

Apply radical rule:

(3 - 1)^{2} = 3 - 1

= 3 - 1

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

Separate the solutions

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Expand

- (- 1 - 3) + 3 - 1 : 23

- (- 1 - 3) + 3 - 1

- (- 1 - 3) : 1 + 3

- (- 1 - 3)

Distribute parentheses

= - (- 1) - (- 3)

Apply minus-plus rules

- (- a) = a

= 1 + 3

= 1 + 3 + 3 - 1

Simplify

1 + 3 + 3 - 1 : 23

1 + 3 + 3 - 1

Add similar elements:

3 + 3 = 23

= 1 + 23 - 1

1 - 1 = 0

= 23

= 23

= \frac{2 3}{2}

Divide the numbers:

\frac{2}{2} = 1

= 3

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Expand

- (- 1 - 3) - (3 - 1) : 2

- (- 1 - 3) - (3 - 1)

- (- 1 - 3) : 1 + 3

- (- 1 - 3)

Distribute parentheses

= - (- 1) - (- 3)

Apply minus-plus rules

- (- a) = a

= 1 + 3

= 1 + 3 - (3 - 1)

- (3 - 1) : - 3 + 1

- (3 - 1)

Distribute parentheses

= - (3) - (- 1)

Apply minus-plus rules

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

= - 3 + 1

= 1 + 3 - 3 + 1

Simplify

1 + 3 - 3 + 1 : 2

1 + 3 - 3 + 1

Add similar elements:

3 - 3 = 0

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

= 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 3, u = 1

Substitute back

u = tan (x)

tan (x) = 3, tan (x) = 1

tan (x) = 3, tan (x) = 1

tan (x) = 3 : x = \frac{π}{3} + πn

tan (x) = 3

General solutions for

tan (x) = 3

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{π}{3} + πn, x = \frac{π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for sec^2(x)=(sqrt(3)+1)tan(x)-sqrt(3)+1 ?
The general solution for sec^2(x)=(sqrt(3)+1)tan(x)-sqrt(3)+1 is x= pi/3+pin,x= pi/4+pin