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=1.20593…+pin,x=0.36486…+pin

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

Solution

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

Solution

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

Degrees

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

= 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}

Apply rule

- (- a) = a

= \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}

Apply rule

- (- a) = a

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

Popular Examples

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sin (θ) + sin (2 θ) = 0

tan^2(θ)sin(θ)=-sin(θ)

tan^{2} (θ) sin (θ) = - sin (θ)

sec(x)=1+tan(x)

sec (x) = 1 + tan (x)

cos^2(x)=0,5

cos^{2} (x) = 0, 5

2cos(x)+3sin(x)=2

2 cos (x) + 3 sin (x) = 2

Frequently Asked Questions (FAQ)

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=1.20593…+pin,x=0.36486…+pin