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

The general solution for 5tan^2(x)-3tan(x)-2=0 is x= pi/4+pin,x=-0.38050…+pin

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

Solution

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

Solution

x = \frac{π}{4} + πn, x = - 0.38050 \dots + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = - 21.80140 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 3)^{2} - 4 \cdot 5 (- 2) = 7

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 5 \cdot 2 = 40

= 3^{2} + 40

3^{2} = 9

= 9 + 40

Add the numbers:

9 + 40 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 3 ) \pm 7}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 7}{2 \cdot 5}, u_{2} = \frac{- ( - 3 ) - 7}{2 \cdot 5}

u = \frac{- ( - 3 ) + 7}{2 \cdot 5} : 1

\frac{- ( - 3 ) + 7}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{3 + 7}{2 \cdot 5}

Add the numbers:

3 + 7 = 10

= \frac{10}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{10}{10}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 3 ) - 7}{2 \cdot 5} : - \frac{2}{5}

\frac{- ( - 3 ) - 7}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{3 - 7}{2 \cdot 5}

Subtract the numbers:

3 - 7 = - 4

= \frac{- 4}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 4}{10}

Apply the fraction rule:

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

= - \frac{4}{10}

Cancel the common factor:

2

= - \frac{2}{5}

The solutions to the quadratic equation are:

u = 1, u = - \frac{2}{5}

Substitute back

u = tan (x)

tan (x) = 1, tan (x) = - \frac{2}{5}

tan (x) = 1, tan (x) = - \frac{2}{5}

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}{5} : x = arctan (- \frac{2}{5}) + πn

tan (x) = - \frac{2}{5}

Apply trig inverse properties

tan (x) = - \frac{2}{5}

General solutions for

tan (x) = - \frac{2}{5}

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

x = arctan (- \frac{2}{5}) + πn

x = arctan (- \frac{2}{5}) + πn

Combine all the solutions

x = \frac{π}{4} + πn, x = arctan (- \frac{2}{5}) + πn

Show solutions in decimal form

x = \frac{π}{4} + πn, x = - 0.38050 \dots + πn

Graph

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solvefor x,tan(x)= 1/2 solve for

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Frequently Asked Questions (FAQ)

What is the general solution for 5tan^2(x)-3tan(x)-2=0 ?
The general solution for 5tan^2(x)-3tan(x)-2=0 is x= pi/4+pin,x=-0.38050…+pin