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

The general solution for tan^2(x)+5tan(x)-7=0 is x=0.85074…+pin,x=-1.40934…+pin

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tan^2(x)+5tan(x)-7=0

tan^{2} (x) + 5 tan (x) - 7 = 0

x = 0.85074 \dots + πn, x = - 1.40934 \dots + πn

Degrees

x = 48.74435 \dots^{\circ} + 18 0^{\circ} n, x = - 80.74974 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) + 5 tan (x) - 7 = 0

Solve by substitution

tan^{2} (x) + 5 tan (x) - 7 = 0

Let:

tan (x) = u

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

u^{2} + 5 u - 7 = 0 : u = \frac{- 5 + 53}{2}, u = \frac{- 5 - 53}{2}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 5, c = - 7

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

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

5^{2} - 4 \cdot 1 \cdot (- 7) = 53

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

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 1 \cdot 7

Multiply the numbers:

4 \cdot 1 \cdot 7 = 28

= 5^{2} + 28

5^{2} = 25

= 25 + 28

Add the numbers:

25 + 28 = 53

= 53

u_{1, 2} = \frac{- 5 \pm 53}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 5 + 53}{2 \cdot 1}, u_{2} = \frac{- 5 - 53}{2 \cdot 1}

u = \frac{- 5 + 53}{2 \cdot 1} : \frac{- 5 + 53}{2}

\frac{- 5 + 53}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 5 + 53}{2}

u = \frac{- 5 - 53}{2 \cdot 1} : \frac{- 5 - 53}{2}

\frac{- 5 - 53}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 5 - 53}{2}

The solutions to the quadratic equation are:

u = \frac{- 5 + 53}{2}, u = \frac{- 5 - 53}{2}

Substitute back

u = tan (x)

tan (x) = \frac{- 5 + 53}{2}, tan (x) = \frac{- 5 - 53}{2}

tan (x) = \frac{- 5 + 53}{2}, tan (x) = \frac{- 5 - 53}{2}

tan (x) = \frac{- 5 + 53}{2} : x = arctan (\frac{- 5 + 53}{2}) + πn

tan (x) = \frac{- 5 + 53}{2}

Apply trig inverse properties

tan (x) = \frac{- 5 + 53}{2}

General solutions for

tan (x) = \frac{- 5 + 53}{2}

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

x = arctan (\frac{- 5 + 53}{2}) + πn

x = arctan (\frac{- 5 + 53}{2}) + πn

tan (x) = \frac{- 5 - 53}{2} : x = arctan (\frac{- 5 - 53}{2}) + πn

tan (x) = \frac{- 5 - 53}{2}

Apply trig inverse properties

tan (x) = \frac{- 5 - 53}{2}

General solutions for

tan (x) = \frac{- 5 - 53}{2}

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

x = arctan (\frac{- 5 - 53}{2}) + πn

x = arctan (\frac{- 5 - 53}{2}) + πn

Combine all the solutions

x = arctan (\frac{- 5 + 53}{2}) + πn, x = arctan (\frac{- 5 - 53}{2}) + πn

Show solutions in decimal form

x = 0.85074 \dots + πn, x = - 1.40934 \dots + πn

sin (x) = \frac{- 3}{4}

cot^{2} (x) - 3 cot (x) = 0

x, y = sec (x)

tan (θ) = 2

cos (a) = 0.5, tan (a) - 3

What is the general solution for tan^2(x)+5tan(x)-7=0 ?
The general solution for tan^2(x)+5tan(x)-7=0 is x=0.85074…+pin,x=-1.40934…+pin