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

The general solution for 2tan^2(x)-tan(x)=0 is x=0.46364…+pin,x=pin

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

Solution

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

Solution

x = 0.46364 \dots + πn, x = πn

Degrees

x = 26.56505 \dots^{\circ} + 18 0^{\circ} n, x = 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

2 u^{2} - u = 0

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

2 u^{2} - u = 0

Solve with the quadratic formula

2 u^{2} - u = 0

Quadratic Equation Formula:

For

a = 2, b = - 1, c = 0

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

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

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

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

4 \cdot 2 \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 2}

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 1 = 0

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = 0

Substitute back

u = tan (x)

tan (x) = \frac{1}{2}, tan (x) = 0

tan (x) = \frac{1}{2}, tan (x) = 0

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

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan (\frac{1}{2}) + πn

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

Show solutions in decimal form

x = 0.46364 \dots + πn, x = πn

Popular Examples

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sqrt(3)tan(x)-2=-1

3 tan (x) - 2 = - 1

solvefor x,1-cos(x)=0 solve for

x, 1 - cos (x) = 0

sin(a)=(-sqrt(3))/2

sin (a) = \frac{- 3}{2}

sin(x)=467

sin (x) = 467

Frequently Asked Questions (FAQ)

What is the general solution for 2tan^2(x)-tan(x)=0 ?
The general solution for 2tan^2(x)-tan(x)=0 is x=0.46364…+pin,x=pin