What is the general solution for 3tan^2(x)-5tan(x)+2=0 ?

The general solution for 3tan^2(x)-5tan(x)+2=0 is x= pi/4+pin,x=0.58800…+pin

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3tan^2(x)-5tan(x)+2=0

Solution

3 tan^{2} (x) - 5 tan (x) + 2 = 0

Solution

x = \frac{π}{4} + πn, x = 0.58800 \dots + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 33.69006 \dots^{\circ} + 18 0^{\circ} n

Solution steps

3 tan^{2} (x) - 5 tan (x) + 2 = 0

Solve by substitution

3 tan^{2} (x) - 5 tan (x) + 2 = 0

Let:

tan (x) = u

3 u^{2} - 5 u + 2 = 0

3 u^{2} - 5 u + 2 = 0 : u = 1, u = \frac{2}{3}

3 u^{2} - 5 u + 2 = 0

Solve with the quadratic formula

3 u^{2} - 5 u + 2 = 0

Quadratic Equation Formula:

For

a = 3, b = - 5, c = 2

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 \cdot 2}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 \cdot 2}{2 \cdot 3}

(- 5)^{2} - 4 \cdot 3 \cdot 2 = 1

(- 5)^{2} - 4 \cdot 3 \cdot 2

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 3 \cdot 2

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 3}, u_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 3}

u = \frac{- ( - 5 ) + 1}{2 \cdot 3} : 1

\frac{- ( - 5 ) + 1}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{5 + 1}{2 \cdot 3}

Add the numbers:

5 + 1 = 6

= \frac{6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 5 ) - 1}{2 \cdot 3} : \frac{2}{3}

\frac{- ( - 5 ) - 1}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{5 - 1}{2 \cdot 3}

Subtract the numbers:

5 - 1 = 4

= \frac{4}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{4}{6}

Cancel the common factor:

2

= \frac{2}{3}

The solutions to the quadratic equation are:

u = 1, u = \frac{2}{3}

Substitute back

u = tan (x)

tan (x) = 1, tan (x) = \frac{2}{3}

tan (x) = 1, tan (x) = \frac{2}{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) = \frac{2}{3} : x = arctan (\frac{2}{3}) + πn

tan (x) = \frac{2}{3}

Apply trig inverse properties

tan (x) = \frac{2}{3}

General solutions for

tan (x) = \frac{2}{3}

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{2}{3}) + πn

x = arctan (\frac{2}{3}) + πn

Combine all the solutions

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

Show solutions in decimal form

x = \frac{π}{4} + πn, x = 0.58800 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3tan^2(x)-5tan(x)+2=0 ?
The general solution for 3tan^2(x)-5tan(x)+2=0 is x= pi/4+pin,x=0.58800…+pin