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

The general solution for tan^2(θ)+5tan(θ)+2=0 is θ=-0.41320…+pin,θ=-1.35498…+pin

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tan^2(θ)+5tan(θ)+2=0

tan^{2} (θ) + 5 tan (θ) + 2 = 0

θ = - 0.41320 \dots + πn, θ = - 1.35498 \dots + πn

Degrees

θ = - 23.67491 \dots^{\circ} + 18 0^{\circ} n, θ = - 77.63501 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) + 5 tan (θ) + 2 = 0

Solve by substitution

tan^{2} (θ) + 5 tan (θ) + 2 = 0

Let:

tan (θ) = u

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

u^{2} + 5 u + 2 = 0 : u = \frac{- 5 + 17}{2}, u = \frac{- 5 - 17}{2}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 5, c = 2

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

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

5^{2} - 4 \cdot 1 \cdot 2 = 17

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 5^{2} - 8

5^{2} = 25

= 25 - 8

Subtract the numbers:

25 - 8 = 17

= 17

u_{1, 2} = \frac{- 5 \pm 17}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 5 + 17}{2 \cdot 1}, u_{2} = \frac{- 5 - 17}{2 \cdot 1}

u = \frac{- 5 + 17}{2 \cdot 1} : \frac{- 5 + 17}{2}

\frac{- 5 + 17}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 5 + 17}{2}

u = \frac{- 5 - 17}{2 \cdot 1} : \frac{- 5 - 17}{2}

\frac{- 5 - 17}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 5 - 17}{2}

The solutions to the quadratic equation are:

u = \frac{- 5 + 17}{2}, u = \frac{- 5 - 17}{2}

Substitute back

u = tan (θ)

tan (θ) = \frac{- 5 + 17}{2}, tan (θ) = \frac{- 5 - 17}{2}

tan (θ) = \frac{- 5 + 17}{2}, tan (θ) = \frac{- 5 - 17}{2}

tan (θ) = \frac{- 5 + 17}{2} : θ = arctan (\frac{- 5 + 17}{2}) + πn

tan (θ) = \frac{- 5 + 17}{2}

Apply trig inverse properties

tan (θ) = \frac{- 5 + 17}{2}

General solutions for

tan (θ) = \frac{- 5 + 17}{2}

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

θ = arctan (\frac{- 5 + 17}{2}) + πn

θ = arctan (\frac{- 5 + 17}{2}) + πn

tan (θ) = \frac{- 5 - 17}{2} : θ = arctan (\frac{- 5 - 17}{2}) + πn

tan (θ) = \frac{- 5 - 17}{2}

Apply trig inverse properties

tan (θ) = \frac{- 5 - 17}{2}

General solutions for

tan (θ) = \frac{- 5 - 17}{2}

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

θ = arctan (\frac{- 5 - 17}{2}) + πn

θ = arctan (\frac{- 5 - 17}{2}) + πn

Combine all the solutions

θ = arctan (\frac{- 5 + 17}{2}) + πn, θ = arctan (\frac{- 5 - 17}{2}) + πn

Show solutions in decimal form

θ = - 0.41320 \dots + πn, θ = - 1.35498 \dots + πn

cos (z) = - 2

sin (2 x) = sin (x), x, \in [0, 2 π]

cos (2 x) + cos (- x) = 0

2 sin^{2} (u) - 3 cos (u) = 0

sin (2 t) = sin (t)

What is the general solution for tan^2(θ)+5tan(θ)+2=0 ?
The general solution for tan^2(θ)+5tan(θ)+2=0 is θ=-0.41320…+pin,θ=-1.35498…+pin