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

The general solution for 5tan^2(x)-tan(x)=0 is x=0.19739…+pin,x=pin

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

Solution

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

Solution

x = 0.19739 \dots + πn, x = πn

Degrees

x = 11.30993 \dots^{\circ} + 18 0^{\circ} n, x = 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

5 u^{2} - u = 0

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

5 u^{2} - u = 0

Solve with the quadratic formula

5 u^{2} - u = 0

Quadratic Equation Formula:

For

a = 5, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

(- 1)^{2} - 4 \cdot 5 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 5 \cdot 0

(- 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

4 \cdot 5 \cdot 0 = 0

4 \cdot 5 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 1 = 2

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{2}{10}

Cancel the common factor:

2

= \frac{1}{5}

u = \frac{- ( - 1 ) - 1}{2 \cdot 5} : 0

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 1 = 0

= \frac{0}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{0}{10}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = \frac{1}{5}, u = 0

Substitute back

u = tan (x)

tan (x) = \frac{1}{5}, tan (x) = 0

tan (x) = \frac{1}{5}, tan (x) = 0

tan (x) = \frac{1}{5} : x = arctan (\frac{1}{5}) + πn

tan (x) = \frac{1}{5}

Apply trig inverse properties

tan (x) = \frac{1}{5}

General solutions for

tan (x) = \frac{1}{5}

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

x = arctan (\frac{1}{5}) + πn

x = arctan (\frac{1}{5}) + πn

tan (x) = 0 : x = πn

tan (x) = 0

General solutions for

tan (x) = 0

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 = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

Combine all the solutions

x = arctan (\frac{1}{5}) + πn, x = πn

Show solutions in decimal form

x = 0.19739 \dots + πn, x = πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5tan^2(x)-tan(x)=0 ?
The general solution for 5tan^2(x)-tan(x)=0 is x=0.19739…+pin,x=pin