What is the general solution for 8tan^2(θ)+10tan(θ)+10=7 ?

The general solution for 8tan^2(θ)+10tan(θ)+10=7 is θ=-0.46364…+pin,θ=-0.64350…+pin

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Popular >

8tan^2(θ)+10tan(θ)+10=7

Solution

8 tan^{2} (θ) + 10 tan (θ) + 10 = 7

Solution

θ = - 0.46364 \dots + πn, θ = - 0.64350 \dots + πn

Degrees

θ = - 26.56505 \dots^{\circ} + 18 0^{\circ} n, θ = - 36.86989 \dots^{\circ} + 18 0^{\circ} n

Solution steps

8 tan^{2} (θ) + 10 tan (θ) + 10 = 7

Solve by substitution

8 tan^{2} (θ) + 10 tan (θ) + 10 = 7

Let:

tan (θ) = u

8 u^{2} + 10 u + 10 = 7

8 u^{2} + 10 u + 10 = 7 : u = - \frac{1}{2}, u = - \frac{3}{4}

8 u^{2} + 10 u + 10 = 7

Move

7

to the left side

8 u^{2} + 10 u + 10 = 7

Subtract

7

from both sides

8 u^{2} + 10 u + 10 - 7 = 7 - 7

Simplify

8 u^{2} + 10 u + 3 = 0

8 u^{2} + 10 u + 3 = 0

Solve with the quadratic formula

8 u^{2} + 10 u + 3 = 0

Quadratic Equation Formula:

For

a = 8, b = 10, c = 3

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

1 0^{2} - 4 \cdot 8 \cdot 3 = 2

1 0^{2} - 4 \cdot 8 \cdot 3

Multiply the numbers:

4 \cdot 8 \cdot 3 = 96

= 1 0^{2} - 96

1 0^{2} = 100

= 100 - 96

Subtract the numbers:

100 - 96 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- 10 \pm 2}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- 10 + 2}{2 \cdot 8}, u_{2} = \frac{- 10 - 2}{2 \cdot 8}

u = \frac{- 10 + 2}{2 \cdot 8} : - \frac{1}{2}

\frac{- 10 + 2}{2 \cdot 8}

Add/Subtract the numbers:

- 10 + 2 = - 8

= \frac{- 8}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 8}{16}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{8}{16}

Cancel the common factor:

8

= - \frac{1}{2}

u = \frac{- 10 - 2}{2 \cdot 8} : - \frac{3}{4}

\frac{- 10 - 2}{2 \cdot 8}

Subtract the numbers:

- 10 - 2 = - 12

= \frac{- 12}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 12}{16}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{12}{16}

Cancel the common factor:

4

= - \frac{3}{4}

The solutions to the quadratic equation are:

u = - \frac{1}{2}, u = - \frac{3}{4}

Substitute back

u = tan (θ)

tan (θ) = - \frac{1}{2}, tan (θ) = - \frac{3}{4}

tan (θ) = - \frac{1}{2}, tan (θ) = - \frac{3}{4}

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{3}{4} : θ = arctan (- \frac{3}{4}) + πn

tan (θ) = - \frac{3}{4}

Apply trig inverse properties

tan (θ) = - \frac{3}{4}

General solutions for

tan (θ) = - \frac{3}{4}

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

θ = arctan (- \frac{3}{4}) + πn

θ = arctan (- \frac{3}{4}) + πn

Combine all the solutions

θ = arctan (- \frac{1}{2}) + πn, θ = arctan (- \frac{3}{4}) + πn

Show solutions in decimal form

θ = - 0.46364 \dots + πn, θ = - 0.64350 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 8tan^2(θ)+10tan(θ)+10=7 ?
The general solution for 8tan^2(θ)+10tan(θ)+10=7 is θ=-0.46364…+pin,θ=-0.64350…+pin