What is the general solution for tan^2(θ)+7tan(θ)+9=0 ?

The general solution for tan^2(θ)+7tan(θ)+9=0 is θ=-1.03835…+pin,θ=-1.38440…+pin

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tan^2(θ)+7tan(θ)+9=0

tan^{2} (θ) + 7 tan (θ) + 9 = 0

θ = - 1.03835 \dots + πn, θ = - 1.38440 \dots + πn

Degrees

θ = - 59.49352 \dots^{\circ} + 18 0^{\circ} n, θ = - 79.32055 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) + 7 tan (θ) + 9 = 0

Solve by substitution

tan^{2} (θ) + 7 tan (θ) + 9 = 0

Let:

tan (θ) = u

u^{2} + 7 u + 9 = 0

u^{2} + 7 u + 9 = 0 : u = \frac{- 7 + 13}{2}, u = \frac{- 7 - 13}{2}

u^{2} + 7 u + 9 = 0

Solve with the quadratic formula

u^{2} + 7 u + 9 = 0

Quadratic Equation Formula:

For

a = 1, b = 7, c = 9

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 9 = 13

7^{2} - 4 \cdot 1 \cdot 9

Multiply the numbers:

4 \cdot 1 \cdot 9 = 36

= 7^{2} - 36

7^{2} = 49

= 49 - 36

Subtract the numbers:

49 - 36 = 13

= 13

u_{1, 2} = \frac{- 7 \pm 13}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 7 + 13}{2 \cdot 1}, u_{2} = \frac{- 7 - 13}{2 \cdot 1}

u = \frac{- 7 + 13}{2 \cdot 1} : \frac{- 7 + 13}{2}

\frac{- 7 + 13}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 7 + 13}{2}

u = \frac{- 7 - 13}{2 \cdot 1} : \frac{- 7 - 13}{2}

\frac{- 7 - 13}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 7 - 13}{2}

The solutions to the quadratic equation are:

u = \frac{- 7 + 13}{2}, u = \frac{- 7 - 13}{2}

Substitute back

u = tan (θ)

tan (θ) = \frac{- 7 + 13}{2}, tan (θ) = \frac{- 7 - 13}{2}

tan (θ) = \frac{- 7 + 13}{2}, tan (θ) = \frac{- 7 - 13}{2}

tan (θ) = \frac{- 7 + 13}{2} : θ = arctan (\frac{- 7 + 13}{2}) + πn

tan (θ) = \frac{- 7 + 13}{2}

Apply trig inverse properties

tan (θ) = \frac{- 7 + 13}{2}

General solutions for

tan (θ) = \frac{- 7 + 13}{2}

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

θ = arctan (\frac{- 7 + 13}{2}) + πn

θ = arctan (\frac{- 7 + 13}{2}) + πn

tan (θ) = \frac{- 7 - 13}{2} : θ = arctan (\frac{- 7 - 13}{2}) + πn

tan (θ) = \frac{- 7 - 13}{2}

Apply trig inverse properties

tan (θ) = \frac{- 7 - 13}{2}

General solutions for

tan (θ) = \frac{- 7 - 13}{2}

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

θ = arctan (\frac{- 7 - 13}{2}) + πn

θ = arctan (\frac{- 7 - 13}{2}) + πn

Combine all the solutions

θ = arctan (\frac{- 7 + 13}{2}) + πn, θ = arctan (\frac{- 7 - 13}{2}) + πn

Show solutions in decimal form

θ = - 1.03835 \dots + πn, θ = - 1.38440 \dots + πn

x, cos (x) \cdot cos (y) = 0

3 sec (θ) = 6

cot (x) = csc (x) + 3

cos (x) \cdot cos (3 5^{\circ}) = 1 - sin (x) \cdot sin (3 5^{\circ})

sin (3 θ - 15) = 0.7

What is the general solution for tan^2(θ)+7tan(θ)+9=0 ?
The general solution for tan^2(θ)+7tan(θ)+9=0 is θ=-1.03835…+pin,θ=-1.38440…+pin