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

The general solution for tan^2(x)-2tan(x)=0 is x=1.10714…+pin,x=pin

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

Solution

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

Solution

x = 1.10714 \dots + πn, x = πn

Degrees

x = 63.43494 \dots^{\circ} + 18 0^{\circ} n, x = 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

u^{2} - 2 u = 0

u^{2} - 2 u = 0 : u = 2, u = 0

u^{2} - 2 u = 0

Solve with the quadratic formula

u^{2} - 2 u = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply exponent rule:

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

n

is even

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

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

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

2 + 2 = 4

= \frac{4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

2 - 2 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = 2, u = 0

Substitute back

u = tan (x)

tan (x) = 2, tan (x) = 0

tan (x) = 2, tan (x) = 0

tan (x) = 2 : x = arctan (2) + πn

tan (x) = 2

Apply trig inverse properties

tan (x) = 2

General solutions for

tan (x) = 2

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

x = arctan (2) + πn

x = arctan (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 (2) + πn, x = πn

Show solutions in decimal form

x = 1.10714 \dots + πn, x = πn

Popular Examples

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sin^{2} (θ) = 1

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sin^{2} (x) - 5 cos (x) + 5 = 0

4sin(x)cos(x)+cos(x)=0

4 sin (x) cos (x) + cos (x) = 0

sin(2x)= 1/(sqrt(2))

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

-2sin^2(x)=3sin(x)+1

- 2 sin^{2} (x) = 3 sin (x) + 1

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)-2tan(x)=0 ?
The general solution for tan^2(x)-2tan(x)=0 is x=1.10714…+pin,x=pin