What is the general solution for tan^2(θ)-tan(θ)=0,θ,0<= θ<2pi ?

Q: What is the general solution for tan^2(θ)-tan(θ)=0,θ,0<= θ<2pi ?

The general solution for tan^2(θ)-tan(θ)=0,θ,0<= θ<2pi is No Solution for θ\in\mathbb{R}

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Popular >

tan^2(θ)-tan(θ)=0,θ,0<= θ<2pi

Solution

tan^{2} (θ) - tan (θ) = 0, θ, 0 \leq θ < 2 π

Solution

No Solution for θ \in R

Solution steps

tan^{2} (θ) - tan (θ) = 0, θ, 0 \leq θ < 2 π

Solve by substitution

tan^{2} (θ) - tan (θ) = 0

Let:

tan (θ) = u

u^{2} - u = 0

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

u^{2} - u = 0

Solve with the quadratic formula

u^{2} - u = 0

Quadratic Equation Formula:

For

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

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

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

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

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

4 \cdot 1 \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 1}

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 1 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 1 = 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 = 1, u = 0

Substitute back

u = tan (θ)

tan (θ) = 1, tan (θ) = 0

tan (θ) = 1, tan (θ) = 0

tan (θ) = 1, θ :

No Solution

tan (θ) = 1, θ

General solutions for

tan (θ) = 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}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Solutions for the range

θ

No Solution

tan (θ) = 0, θ :

No Solution

tan (θ) = 0, θ

General solutions for

tan (θ) = 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}

θ = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

Solutions for the range

θ

No Solution

Combine all the solutions

No Solution for θ \in R

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cos (4 x) - 4 cos (2 x) = 5

sin(x)=sin(x+(-3)pi)

sin (x) = sin (x + (- 3) π)

sin(2x)-2/3 =0

sin (2 x) - \frac{2}{3} = 0

2sin(θ)+cos(2θ)=0

2 sin (θ) + cos (2 θ) = 0

sin(x+pi/4)=sqrt(2)cos(x+pi/4)

sin (x + \frac{π}{4}) = 2 cos (x + \frac{π}{4})

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)-tan(θ)=0,θ,0<= θ<2pi ?
The general solution for tan^2(θ)-tan(θ)=0,θ,0<= θ<2pi is No Solution for θ\in\mathbb{R}