What is the general solution for 8tan^2(θ)-1=3tan(θ)+4 ?

The general solution for 8tan^2(θ)-1=3tan(θ)+4 is θ= pi/4+pin,θ=-0.55859…+pin

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Popular Trigonometry >

8tan^2(θ)-1=3tan(θ)+4

Solution

8 tan^{2} (θ) - 1 = 3 tan (θ) + 4

Solution

θ = \frac{π}{4} + πn, θ = - 0.55859 \dots + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = - 32.00538 \dots^{\circ} + 18 0^{\circ} n

Solution steps

8 tan^{2} (θ) - 1 = 3 tan (θ) + 4

Solve by substitution

8 tan^{2} (θ) - 1 = 3 tan (θ) + 4

Let:

tan (θ) = u

8 u^{2} - 1 = 3 u + 4

8 u^{2} - 1 = 3 u + 4 : u = 1, u = - \frac{5}{8}

8 u^{2} - 1 = 3 u + 4

Move

4

to the left side

8 u^{2} - 1 = 3 u + 4

Subtract

4

from both sides

8 u^{2} - 1 - 4 = 3 u + 4 - 4

Simplify

8 u^{2} - 5 = 3 u

8 u^{2} - 5 = 3 u

Move

3 u

to the left side

8 u^{2} - 5 = 3 u

Subtract

3 u

from both sides

8 u^{2} - 5 - 3 u = 3 u - 3 u

Simplify

8 u^{2} - 5 - 3 u = 0

8 u^{2} - 5 - 3 u = 0

Write in the standard form

a x^{2} + b x + c = 0

8 u^{2} - 3 u - 5 = 0

Solve with the quadratic formula

8 u^{2} - 3 u - 5 = 0

Quadratic Equation Formula:

For

a = 8, b = - 3, c = - 5

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 8 ( - 5 )}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 8 ( - 5 )}{2 \cdot 8}

(- 3)^{2} - 4 \cdot 8 (- 5) = 13

(- 3)^{2} - 4 \cdot 8 (- 5)

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 8 \cdot 5

Apply exponent rule:

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

n

is even

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

= 3^{2} + 4 \cdot 8 \cdot 5

Multiply the numbers:

4 \cdot 8 \cdot 5 = 160

= 3^{2} + 160

3^{2} = 9

= 9 + 160

Add the numbers:

9 + 160 = 169

= 169

Factor the number:

169 = 1 3^{2}

= 1 3^{2}

Apply radical rule:

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- ( - 3 ) \pm 13}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 13}{2 \cdot 8}, u_{2} = \frac{- ( - 3 ) - 13}{2 \cdot 8}

u = \frac{- ( - 3 ) + 13}{2 \cdot 8} : 1

\frac{- ( - 3 ) + 13}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{3 + 13}{2 \cdot 8}

Add the numbers:

3 + 13 = 16

= \frac{16}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{16}{16}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 3 ) - 13}{2 \cdot 8} : - \frac{5}{8}

\frac{- ( - 3 ) - 13}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{3 - 13}{2 \cdot 8}

Subtract the numbers:

3 - 13 = - 10

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

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 10}{16}

Apply the fraction rule:

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

= - \frac{10}{16}

Cancel the common factor:

2

= - \frac{5}{8}

The solutions to the quadratic equation are:

u = 1, u = - \frac{5}{8}

Substitute back

u = tan (θ)

tan (θ) = 1, tan (θ) = - \frac{5}{8}

tan (θ) = 1, tan (θ) = - \frac{5}{8}

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

tan (θ) = - \frac{5}{8} : θ = arctan (- \frac{5}{8}) + πn

tan (θ) = - \frac{5}{8}

Apply trig inverse properties

tan (θ) = - \frac{5}{8}

General solutions for

tan (θ) = - \frac{5}{8}

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

θ = arctan (- \frac{5}{8}) + πn

θ = arctan (- \frac{5}{8}) + πn

Combine all the solutions

θ = \frac{π}{4} + πn, θ = arctan (- \frac{5}{8}) + πn

Show solutions in decimal form

θ = \frac{π}{4} + πn, θ = - 0.55859 \dots + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 8tan^2(θ)-1=3tan(θ)+4 ?
The general solution for 8tan^2(θ)-1=3tan(θ)+4 is θ= pi/4+pin,θ=-0.55859…+pin