What is the general solution for -5tan^2(θ)+3=8tan(θ)+6 ?

The general solution for -5tan^2(θ)+3=8tan(θ)+6 is θ=(3pi)/4+pin,θ=-0.54041…+pin

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Popular >

-5tan^2(θ)+3=8tan(θ)+6

Solution

- 5 tan^{2} (θ) + 3 = 8 tan (θ) + 6

Solution

θ = \frac{3 π}{4} + πn, θ = - 0.54041 \dots + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n, θ = - 30.96375 \dots^{\circ} + 18 0^{\circ} n

Solution steps

- 5 tan^{2} (θ) + 3 = 8 tan (θ) + 6

Solve by substitution

- 5 tan^{2} (θ) + 3 = 8 tan (θ) + 6

Let:

tan (θ) = u

- 5 u^{2} + 3 = 8 u + 6

- 5 u^{2} + 3 = 8 u + 6 : u = - 1, u = - \frac{3}{5}

- 5 u^{2} + 3 = 8 u + 6

Move

6

to the left side

- 5 u^{2} + 3 = 8 u + 6

Subtract

6

from both sides

- 5 u^{2} + 3 - 6 = 8 u + 6 - 6

Simplify

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

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

Move

8 u

to the left side

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

Subtract

8 u

from both sides

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

Simplify

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

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

Multiply the numbers:

4 \cdot 5 \cdot 3 = 60

= 8^{2} - 60

8^{2} = 64

= 64 - 60

Subtract the numbers:

64 - 60 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- ( - 8 ) \pm 2}{2 ( - 5 )}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 2}{2 ( - 5 )}, u_{2} = \frac{- ( - 8 ) - 2}{2 ( - 5 )}

u = \frac{- ( - 8 ) + 2}{2 ( - 5 )} : - 1

\frac{- ( - 8 ) + 2}{2 ( - 5 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{8 + 2}{- 2 \cdot 5}

Add the numbers:

8 + 2 = 10

= \frac{10}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{10}{- 10}

Apply the fraction rule:

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

= - \frac{10}{10}

Apply rule

\frac{a}{a} = 1

= - 1

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

\frac{- ( - 8 ) - 2}{2 ( - 5 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{8 - 2}{- 2 \cdot 5}

Subtract the numbers:

8 - 2 = 6

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{6}{- 10}

Apply the fraction rule:

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

= - \frac{6}{10}

Cancel the common factor:

2

= - \frac{3}{5}

The solutions to the quadratic equation are:

u = - 1, u = - \frac{3}{5}

Substitute back

u = tan (θ)

tan (θ) = - 1, tan (θ) = - \frac{3}{5}

tan (θ) = - 1, tan (θ) = - \frac{3}{5}

tan (θ) = - 1 : θ = \frac{3 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{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

tan (θ) = - \frac{3}{5} : θ = arctan (- \frac{3}{5}) + πn

tan (θ) = - \frac{3}{5}

Apply trig inverse properties

tan (θ) = - \frac{3}{5}

General solutions for

tan (θ) = - \frac{3}{5}

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

θ = arctan (- \frac{3}{5}) + πn

θ = arctan (- \frac{3}{5}) + πn

Combine all the solutions

θ = \frac{3 π}{4} + πn, θ = arctan (- \frac{3}{5}) + πn

Show solutions in decimal form

θ = \frac{3 π}{4} + πn, θ = - 0.54041 \dots + πn

Popular Examples

cos(θ)=(-3)/(sqrt(13))

cos (θ) = \frac{- 3}{13}

cos(x)=0.84

cos (x) = 0.84

cos(2x)=((4k-5))/(10)

cos (2 x) = \frac{( 4 k - 5 )}{10}

cos^2(2x)-sin^2(2x)=-1

cos^{2} (2 x) - sin^{2} (2 x) = - 1

cos(x+pi/4)=1

cos (x + \frac{π}{4}) = 1

Frequently Asked Questions (FAQ)

What is the general solution for -5tan^2(θ)+3=8tan(θ)+6 ?
The general solution for -5tan^2(θ)+3=8tan(θ)+6 is θ=(3pi)/4+pin,θ=-0.54041…+pin