What is the general solution for 6tan^2(θ)-10tan(θ)+1=-5tan(θ) ?

The general solution for 6tan^2(θ)-10tan(θ)+1=-5tan(θ) is θ=0.46364…+pin,θ=0.32175…+pin

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Popular >

6tan^2(θ)-10tan(θ)+1=-5tan(θ)

Solution

6 tan^{2} (θ) - 10 tan (θ) + 1 = - 5 tan (θ)

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) - 10 tan (θ) + 1 = - 5 tan (θ)

Solve by substitution

6 tan^{2} (θ) - 10 tan (θ) + 1 = - 5 tan (θ)

Let:

tan (θ) = u

6 u^{2} - 10 u + 1 = - 5 u

6 u^{2} - 10 u + 1 = - 5 u : u = \frac{1}{2}, u = \frac{1}{3}

6 u^{2} - 10 u + 1 = - 5 u

Move

5 u

to the left side

6 u^{2} - 10 u + 1 = - 5 u

Add

5 u

to both sides

6 u^{2} - 10 u + 1 + 5 u = - 5 u + 5 u

Simplify

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

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

Popular Examples

2sin(2x)+6sin(x)-2cos(x)=3

2 sin (2 x) + 6 sin (x) - 2 cos (x) = 3

sin(2x)= 4/5

sin (2 x) = \frac{4}{5}

2csc(x)+17=15+csc(x)

2 csc (x) + 17 = 15 + csc (x)

cos(x)-sin(x)=sqrt(2)

cos (x) - sin (x) = 2

cot(θ)=(5.8473)

cot (θ) = (5.8473)

Frequently Asked Questions (FAQ)

What is the general solution for 6tan^2(θ)-10tan(θ)+1=-5tan(θ) ?
The general solution for 6tan^2(θ)-10tan(θ)+1=-5tan(θ) is θ=0.46364…+pin,θ=0.32175…+pin