What is the general solution for tan^2(x)+3tan(x)+1=0 ?

The general solution for tan^2(x)+3tan(x)+1=0 is x=-0.36486…+pin,x=-1.20593…+pin

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tan^2(x)+3tan(x)+1=0

tan^{2} (x) + 3 tan (x) + 1 = 0

x = - 0.36486 \dots + πn, x = - 1.20593 \dots + πn

Degrees

x = - 20.90515 \dots^{\circ} + 18 0^{\circ} n, x = - 69.09484 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) + 3 tan (x) + 1 = 0

Solve by substitution

tan^{2} (x) + 3 tan (x) + 1 = 0

Let:

tan (x) = u

u^{2} + 3 u + 1 = 0

u^{2} + 3 u + 1 = 0 : u = \frac{- 3 + 5}{2}, u = \frac{- 3 - 5}{2}

u^{2} + 3 u + 1 = 0

Solve with the quadratic formula

u^{2} + 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 3, c = 1

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot 1 = 5

3^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 3^{2} - 4

3^{2} = 9

= 9 - 4

Subtract the numbers:

9 - 4 = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 3 + 5}{2 \cdot 1}, u_{2} = \frac{- 3 - 5}{2 \cdot 1}

u = \frac{- 3 + 5}{2 \cdot 1} : \frac{- 3 + 5}{2}

\frac{- 3 + 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 3 + 5}{2}

u = \frac{- 3 - 5}{2 \cdot 1} : \frac{- 3 - 5}{2}

\frac{- 3 - 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 3 - 5}{2}

The solutions to the quadratic equation are:

u = \frac{- 3 + 5}{2}, u = \frac{- 3 - 5}{2}

Substitute back

u = tan (x)

tan (x) = \frac{- 3 + 5}{2}, tan (x) = \frac{- 3 - 5}{2}

tan (x) = \frac{- 3 + 5}{2}, tan (x) = \frac{- 3 - 5}{2}

tan (x) = \frac{- 3 + 5}{2} : x = arctan (\frac{- 3 + 5}{2}) + πn

tan (x) = \frac{- 3 + 5}{2}

Apply trig inverse properties

tan (x) = \frac{- 3 + 5}{2}

General solutions for

tan (x) = \frac{- 3 + 5}{2}

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

x = arctan (\frac{- 3 + 5}{2}) + πn

x = arctan (\frac{- 3 + 5}{2}) + πn

tan (x) = \frac{- 3 - 5}{2} : x = arctan (\frac{- 3 - 5}{2}) + πn

tan (x) = \frac{- 3 - 5}{2}

Apply trig inverse properties

tan (x) = \frac{- 3 - 5}{2}

General solutions for

tan (x) = \frac{- 3 - 5}{2}

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

x = arctan (\frac{- 3 - 5}{2}) + πn

x = arctan (\frac{- 3 - 5}{2}) + πn

Combine all the solutions

x = arctan (\frac{- 3 + 5}{2}) + πn, x = arctan (\frac{- 3 - 5}{2}) + πn

Show solutions in decimal form

x = - 0.36486 \dots + πn, x = - 1.20593 \dots + πn

csc (2 x) = 2, 0 \leq x < 2 π

sin (x) + cos (x) = 1.2

30 cos (30 x) + 14 = - 16

x, cot (- x) cos (- x) + sin (- x) = - \frac{1}{f}

cos (x) = - \frac{2}{3}

What is the general solution for tan^2(x)+3tan(x)+1=0 ?
The general solution for tan^2(x)+3tan(x)+1=0 is x=-0.36486…+pin,x=-1.20593…+pin