What is the general solution for tan^2(θ)-6tan(θ)+5=0 ?

The general solution for tan^2(θ)-6tan(θ)+5=0 is θ=1.37340…+pin,θ= pi/4+pin

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tan^2(θ)-6tan(θ)+5=0

Solution

tan^{2} (θ) - 6 tan (θ) + 5 = 0

Solution

θ = 1.37340 \dots + πn, θ = \frac{π}{4} + πn

Degrees

θ = 78.69006 \dots^{\circ} + 18 0^{\circ} n, θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) - 6 tan (θ) + 5 = 0

Solve by substitution

tan^{2} (θ) - 6 tan (θ) + 5 = 0

Let:

tan (θ) = u

u^{2} - 6 u + 5 = 0

u^{2} - 6 u + 5 = 0 : u = 5, u = 1

u^{2} - 6 u + 5 = 0

Solve with the quadratic formula

u^{2} - 6 u + 5 = 0

Quadratic Equation Formula:

For

a = 1, b = - 6, c = 5

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 5 = 4

(- 6)^{2} - 4 \cdot 1 \cdot 5

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 1 \cdot 5

Multiply the numbers:

4 \cdot 1 \cdot 5 = 20

= 6^{2} - 20

6^{2} = 36

= 36 - 20

Subtract the numbers:

36 - 20 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

u_{1, 2} = \frac{- ( - 6 ) \pm 4}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 4}{2 \cdot 1}, u_{2} = \frac{- ( - 6 ) - 4}{2 \cdot 1}

u = \frac{- ( - 6 ) + 4}{2 \cdot 1} : 5

\frac{- ( - 6 ) + 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 + 4}{2 \cdot 1}

Add the numbers:

6 + 4 = 10

= \frac{10}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{10}{2}

Divide the numbers:

\frac{10}{2} = 5

= 5

u = \frac{- ( - 6 ) - 4}{2 \cdot 1} : 1

\frac{- ( - 6 ) - 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 - 4}{2 \cdot 1}

Subtract the numbers:

6 - 4 = 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 = 5, u = 1

Substitute back

u = tan (θ)

tan (θ) = 5, tan (θ) = 1

tan (θ) = 5, tan (θ) = 1

tan (θ) = 5 : θ = arctan (5) + πn

tan (θ) = 5

Apply trig inverse properties

tan (θ) = 5

General solutions for

tan (θ) = 5

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

θ = arctan (5) + πn

θ = arctan (5) + 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 (5) + πn, θ = \frac{π}{4} + πn

Show solutions in decimal form

θ = 1.37340 \dots + πn, θ = \frac{π}{4} + πn

Popular Examples

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tan(x)=2.5

tan (x) = 2.5

solvefor y,arctan(y)=x solve for

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)-6tan(θ)+5=0 ?
The general solution for tan^2(θ)-6tan(θ)+5=0 is θ=1.37340…+pin,θ= pi/4+pin