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

The general solution for tan^2(θ)-3tan(θ)+2=0 is θ=1.10714…+pin,θ= pi/4+pin

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tan^2(θ)-3tan(θ)+2=0

Solution

tan^{2} (θ) - 3 tan (θ) + 2 = 0

Solution

θ = 1.10714 \dots + πn, θ = \frac{π}{4} + πn

Degrees

θ = 63.43494 \dots^{\circ} + 18 0^{\circ} n, θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) - 3 tan (θ) + 2 = 0

Solve by substitution

tan^{2} (θ) - 3 tan (θ) + 2 = 0

Let:

tan (θ) = u

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

u^{2} - 3 u + 2 = 0 : u = 2, u = 1

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = - 3, c = 2

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

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

(- 3)^{2} - 4 \cdot 1 \cdot 2 = 1

(- 3)^{2} - 4 \cdot 1 \cdot 2

Apply exponent rule:

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

n

is even

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

= 3^{2} - 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

\frac{- ( - 3 ) + 1}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 + 1}{2 \cdot 1}

Add the numbers:

3 + 1 = 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{- ( - 3 ) - 1}{2 \cdot 1} : 1

\frac{- ( - 3 ) - 1}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 - 1}{2 \cdot 1}

Subtract the numbers:

3 - 1 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 2, u = 1

Substitute back

u = tan (θ)

tan (θ) = 2, tan (θ) = 1

tan (θ) = 2, tan (θ) = 1

tan (θ) = 2 : θ = arctan (2) + πn

tan (θ) = 2

Apply trig inverse properties

tan (θ) = 2

General solutions for

tan (θ) = 2

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

θ = arctan (2) + πn

θ = arctan (2) + πn

tan (θ) = 1 : θ = \frac{π}{4} + πn

tan (θ) = 1

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

Combine all the solutions

θ = arctan (2) + πn, θ = \frac{π}{4} + πn

Show solutions in decimal form

θ = 1.10714 \dots + πn, θ = \frac{π}{4} + πn

Popular Examples

tan(θ)= 1/(sqrt(3))

tan (θ) = \frac{1}{3}

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

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

4sin(x)cos(x)=1

4 sin (x) cos (x) = 1

sec^2(x)-sec(x)=2

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

sin(x)= 4/5

sin (x) = \frac{4}{5}

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)-3tan(θ)+2=0 ?
The general solution for tan^2(θ)-3tan(θ)+2=0 is θ=1.10714…+pin,θ= pi/4+pin