What is the general solution for 6tan^2(θ)-5tan(θ)+1=0 ?

The general solution for 6tan^2(θ)-5tan(θ)+1=0 is θ=0.46364…+pin,θ=0.32175…+pin

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Popular >

6tan^2(θ)-5tan(θ)+1=0

Solution

6 tan^{2} (θ) - 5 tan (θ) + 1 = 0

Solution

θ = 0.46364 \dots + πn, θ = 0.32175 \dots + πn

Degrees

θ = 26.56505 \dots^{\circ} + 18 0^{\circ} n, θ = 18.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

6 tan^{2} (θ) - 5 tan (θ) + 1 = 0

Solve by substitution

6 tan^{2} (θ) - 5 tan (θ) + 1 = 0

Let:

tan (θ) = u

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

6 u^{2} - 5 u + 1 = 0 : u = \frac{1}{2}, u = \frac{1}{3}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

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

Multiply the numbers:

4 \cdot 6 \cdot 1 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 6}, u_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 6}

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

\frac{- ( - 5 ) + 1}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{5 + 1}{2 \cdot 6}

Add the numbers:

5 + 1 = 6

= \frac{6}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{6}{12}

Cancel the common factor:

6

= \frac{1}{2}

u = \frac{- ( - 5 ) - 1}{2 \cdot 6} : \frac{1}{3}

\frac{- ( - 5 ) - 1}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{5 - 1}{2 \cdot 6}

Subtract the numbers:

5 - 1 = 4

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{4}{12}

Cancel the common factor:

4

= \frac{1}{3}

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = \frac{1}{3}

Substitute back

u = tan (θ)

tan (θ) = \frac{1}{2}, tan (θ) = \frac{1}{3}

tan (θ) = \frac{1}{2}, tan (θ) = \frac{1}{3}

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

tan (θ) = \frac{1}{2}

Apply trig inverse properties

tan (θ) = \frac{1}{2}

General solutions for

tan (θ) = \frac{1}{2}

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

θ = arctan (\frac{1}{2}) + πn

θ = arctan (\frac{1}{2}) + πn

tan (θ) = \frac{1}{3} : θ = arctan (\frac{1}{3}) + πn

tan (θ) = \frac{1}{3}

Apply trig inverse properties

tan (θ) = \frac{1}{3}

General solutions for

tan (θ) = \frac{1}{3}

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

θ = arctan (\frac{1}{3}) + πn

θ = arctan (\frac{1}{3}) + πn

Combine all the solutions

θ = arctan (\frac{1}{2}) + πn, θ = arctan (\frac{1}{3}) + πn

Show solutions in decimal form

θ = 0.46364 \dots + πn, θ = 0.32175 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 6tan^2(θ)-5tan(θ)+1=0 ?
The general solution for 6tan^2(θ)-5tan(θ)+1=0 is θ=0.46364…+pin,θ=0.32175…+pin