What is the general solution for tan^2(θ)+7tan(θ)+5=0 ?

The general solution for tan^2(θ)+7tan(θ)+5=0 is θ=-0.67924…+pin,θ=-1.41069…+pin

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tan^2(θ)+7tan(θ)+5=0

tan^{2} (θ) + 7 tan (θ) + 5 = 0

θ = - 0.67924 \dots + πn, θ = - 1.41069 \dots + πn

Degrees

θ = - 38.91801 \dots^{\circ} + 18 0^{\circ} n, θ = - 80.82686 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) + 7 tan (θ) + 5 = 0

Solve by substitution

tan^{2} (θ) + 7 tan (θ) + 5 = 0

Let:

tan (θ) = u

u^{2} + 7 u + 5 = 0

u^{2} + 7 u + 5 = 0 : u = \frac{- 7 + 29}{2}, u = \frac{- 7 - 29}{2}

u^{2} + 7 u + 5 = 0

Solve with the quadratic formula

u^{2} + 7 u + 5 = 0

Quadratic Equation Formula:

For

a = 1, b = 7, c = 5

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 5 = 29

7^{2} - 4 \cdot 1 \cdot 5

Multiply the numbers:

4 \cdot 1 \cdot 5 = 20

= 7^{2} - 20

7^{2} = 49

= 49 - 20

Subtract the numbers:

49 - 20 = 29

= 29

u_{1, 2} = \frac{- 7 \pm 29}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 7 + 29}{2 \cdot 1}, u_{2} = \frac{- 7 - 29}{2 \cdot 1}

u = \frac{- 7 + 29}{2 \cdot 1} : \frac{- 7 + 29}{2}

\frac{- 7 + 29}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 7 + 29}{2}

u = \frac{- 7 - 29}{2 \cdot 1} : \frac{- 7 - 29}{2}

\frac{- 7 - 29}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 7 - 29}{2}

The solutions to the quadratic equation are:

u = \frac{- 7 + 29}{2}, u = \frac{- 7 - 29}{2}

Substitute back

u = tan (θ)

tan (θ) = \frac{- 7 + 29}{2}, tan (θ) = \frac{- 7 - 29}{2}

tan (θ) = \frac{- 7 + 29}{2}, tan (θ) = \frac{- 7 - 29}{2}

tan (θ) = \frac{- 7 + 29}{2} : θ = arctan (\frac{- 7 + 29}{2}) + πn

tan (θ) = \frac{- 7 + 29}{2}

Apply trig inverse properties

tan (θ) = \frac{- 7 + 29}{2}

General solutions for

tan (θ) = \frac{- 7 + 29}{2}

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

θ = arctan (\frac{- 7 + 29}{2}) + πn

θ = arctan (\frac{- 7 + 29}{2}) + πn

tan (θ) = \frac{- 7 - 29}{2} : θ = arctan (\frac{- 7 - 29}{2}) + πn

tan (θ) = \frac{- 7 - 29}{2}

Apply trig inverse properties

tan (θ) = \frac{- 7 - 29}{2}

General solutions for

tan (θ) = \frac{- 7 - 29}{2}

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

θ = arctan (\frac{- 7 - 29}{2}) + πn

θ = arctan (\frac{- 7 - 29}{2}) + πn

Combine all the solutions

θ = arctan (\frac{- 7 + 29}{2}) + πn, θ = arctan (\frac{- 7 - 29}{2}) + πn

Show solutions in decimal form

θ = - 0.67924 \dots + πn, θ = - 1.41069 \dots + πn

tan (θ) = \frac{2 3}{1}

cos^{2} (x) - cos (x) = 0, x, 0 \leq x \leq 2 π

6 cos^{2} (x) + sin (x) - 4 = 0

2 sin (4 x) - 3 = 0

cos (4 x) = \frac{1}{2}, 0 \leq x \leq 2 π

What is the general solution for tan^2(θ)+7tan(θ)+5=0 ?
The general solution for tan^2(θ)+7tan(θ)+5=0 is θ=-0.67924…+pin,θ=-1.41069…+pin