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

The general solution for 3tan^2(x)-2sqrt(3)tan(x)+1=0 is x= pi/6+pin

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

Solution

3 tan^{2} (x) - 23 tan (x) + 1 = 0

Solution

x = \frac{π}{6} + πn

Degrees

x = 3 0^{\circ} + 18 0^{\circ} n

Solution steps

3 tan^{2} (x) - 23 tan (x) + 1 = 0

Solve by substitution

3 tan^{2} (x) - 23 tan (x) + 1 = 0

Let:

tan (x) = u

3 u^{2} - 23 u + 1 = 0

3 u^{2} - 23 u + 1 = 0 : u = \frac{3}{3}

3 u^{2} - 23 u + 1 = 0

Solve with the quadratic formula

3 u^{2} - 23 u + 1 = 0

Quadratic Equation Formula:

For

a = 3, b = - 23, c = 1

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

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

(- 23)^{2} - 4 \cdot 3 \cdot 1 = 0

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

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

(- 23)^{2}

Apply exponent rule:

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

n

is even

(- 23)^{2} = (23)^{2}

= (23)^{2}

Apply exponent rule:

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

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

(3)^{2} : 3

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

= 2^{2} \cdot 3

4 \cdot 3 \cdot 1 = 12

4 \cdot 3 \cdot 1

Multiply the numbers:

4 \cdot 3 \cdot 1 = 12

= 12

= 2^{2} \cdot 3 - 12

2^{2} \cdot 3 = 12

2^{2} \cdot 3

2^{2} = 4

= 4 \cdot 3

Multiply the numbers:

4 \cdot 3 = 12

= 12

= 12 - 12

Subtract the numbers:

12 - 12 = 0

= 0

u_{1, 2} = \frac{- ( - 2 3 ) \pm 0}{2 \cdot 3}

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

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

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

Apply rule

- (- a) = a

= \frac{2 3}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{2 3}{6}

Cancel the common factor:

2

= \frac{3}{3}

u = \frac{3}{3}

The solution to the quadratic equation is:

u = \frac{3}{3}

Substitute back

u = tan (x)

tan (x) = \frac{3}{3}

tan (x) = \frac{3}{3}

tan (x) = \frac{3}{3} : x = \frac{π}{6} + πn

tan (x) = \frac{3}{3}

General solutions for

tan (x) = \frac{3}{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}{6} + πn

x = \frac{π}{6} + πn

Combine all the solutions

x = \frac{π}{6} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3tan^2(x)-2sqrt(3)tan(x)+1=0 ?
The general solution for 3tan^2(x)-2sqrt(3)tan(x)+1=0 is x= pi/6+pin