What is the general solution for tan^2(x)+5tan(x)-1=0 ?

The general solution for tan^2(x)+5tan(x)-1=0 is x=0.19025…+pin,x=-1.38054…+pin

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tan^2(x)+5tan(x)-1=0

tan^{2} (x) + 5 tan (x) - 1 = 0

x = 0.19025 \dots + πn, x = - 1.38054 \dots + πn

Degrees

x = 10.90070 \dots^{\circ} + 18 0^{\circ} n, x = - 79.09929 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) + 5 tan (x) - 1 = 0

Solve by substitution

tan^{2} (x) + 5 tan (x) - 1 = 0

Let:

tan (x) = u

u^{2} + 5 u - 1 = 0

u^{2} + 5 u - 1 = 0 : u = \frac{- 5 + 29}{2}, u = \frac{- 5 - 29}{2}

u^{2} + 5 u - 1 = 0

Solve with the quadratic formula

u^{2} + 5 u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 5, c = - 1

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

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

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

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

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 5^{2} + 4

5^{2} = 25

= 25 + 4

Add the numbers:

25 + 4 = 29

= 29

u_{1, 2} = \frac{- 5 \pm 29}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 5 + 29}{2 \cdot 1}, u_{2} = \frac{- 5 - 29}{2 \cdot 1}

u = \frac{- 5 + 29}{2 \cdot 1} : \frac{- 5 + 29}{2}

\frac{- 5 + 29}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 5 + 29}{2}

u = \frac{- 5 - 29}{2 \cdot 1} : \frac{- 5 - 29}{2}

\frac{- 5 - 29}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 5 - 29}{2}

The solutions to the quadratic equation are:

u = \frac{- 5 + 29}{2}, u = \frac{- 5 - 29}{2}

Substitute back

u = tan (x)

tan (x) = \frac{- 5 + 29}{2}, tan (x) = \frac{- 5 - 29}{2}

tan (x) = \frac{- 5 + 29}{2}, tan (x) = \frac{- 5 - 29}{2}

tan (x) = \frac{- 5 + 29}{2} : x = arctan (\frac{- 5 + 29}{2}) + πn

tan (x) = \frac{- 5 + 29}{2}

Apply trig inverse properties

tan (x) = \frac{- 5 + 29}{2}

General solutions for

tan (x) = \frac{- 5 + 29}{2}

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

x = arctan (\frac{- 5 + 29}{2}) + πn

x = arctan (\frac{- 5 + 29}{2}) + πn

tan (x) = \frac{- 5 - 29}{2} : x = arctan (\frac{- 5 - 29}{2}) + πn

tan (x) = \frac{- 5 - 29}{2}

Apply trig inverse properties

tan (x) = \frac{- 5 - 29}{2}

General solutions for

tan (x) = \frac{- 5 - 29}{2}

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

x = arctan (\frac{- 5 - 29}{2}) + πn

x = arctan (\frac{- 5 - 29}{2}) + πn

Combine all the solutions

x = arctan (\frac{- 5 + 29}{2}) + πn, x = arctan (\frac{- 5 - 29}{2}) + πn

Show solutions in decimal form

x = 0.19025 \dots + πn, x = - 1.38054 \dots + πn

csc (θ) + 1 = 0

2 sin^{2} (θ) - 3 sin (θ) + 1 = 0, 0 \leq θ \leq 2 π

2 sec (x) + 2 tan (x) = 2

2 sin^{2} (θ) + 3 sin (θ) = 0

cos^{2} (x) = sin^{2} (\frac{x}{2})

What is the general solution for tan^2(x)+5tan(x)-1=0 ?
The general solution for tan^2(x)+5tan(x)-1=0 is x=0.19025…+pin,x=-1.38054…+pin