What is the general solution for tan^2(x)+(sqrt(3)-1)tan(x)-sqrt(3)=0 ?

The general solution for tan^2(x)+(sqrt(3)-1)tan(x)-sqrt(3)=0 is x= pi/4+pin,x=(2pi)/3+pin

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Popular Trigonometry >

tan^2(x)+(sqrt(3)-1)tan(x)-sqrt(3)=0

Solution

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

Solution

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

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 = 4

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

Expand

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

(3 - 1)^{2} + 43

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

Apply Perfect Square Formula:

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

a = 3, b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

(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

23 \cdot 1 = 23

23 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 23

= 3 - 23 + 1

Add the numbers:

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

Separate the solutions

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Expand

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

- (3 - 1) + 3 + 1

- (3 - 1) : - 3 + 1

- (3 - 1)

Distribute parentheses

= - (3) - (- 1)

Apply minus-plus rules

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

= - 3 + 1

= - 3 + 1 + 3 + 1

Simplify

- 3 + 1 + 3 + 1 : 2

- 3 + 1 + 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

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Expand

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

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

- (3 - 1) : - 3 + 1

- (3 - 1)

Distribute parentheses

= - (3) - (- 1)

Apply minus-plus rules

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

= - 3 + 1

= - 3 + 1 - (3 + 1)

- (3 + 1) : - 3 - 1

- (3 + 1)

Distribute parentheses

= - (3) - (1)

Apply minus-plus rules

+ (- a) = - a

= - 3 - 1

= - 3 + 1 - 3 - 1

Simplify

- 3 + 1 - 3 - 1 : - 23

- 3 + 1 - 3 - 1

Add similar elements:

- 3 - 3 = - 23

= - 23 + 1 - 1

1 - 1 = 0

= - 23

= - 23

= \frac{- 2 3}{2}

Apply the fraction rule:

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

= - \frac{2 3}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 3

The solutions to the quadratic equation are:

u = 1, u = - 3

Substitute back

u = tan (x)

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

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

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

tan (x) = - 3 : x = \frac{2 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{2 π}{3} + πn

x = \frac{2 π}{3} + πn

Combine all the solutions

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)+(sqrt(3)-1)tan(x)-sqrt(3)=0 ?
The general solution for tan^2(x)+(sqrt(3)-1)tan(x)-sqrt(3)=0 is x= pi/4+pin,x=(2pi)/3+pin