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

The general solution for 2tan^2(θ)-3tan(θ)+1=0 is θ= pi/4+pin,θ=0.46364…+pin

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2tan^2(θ)-3tan(θ)+1=0

Solution

2 tan^{2} (θ) - 3 tan (θ) + 1 = 0

Solution

θ = \frac{π}{4} + πn, θ = 0.46364 \dots + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

2 tan^{2} (θ) - 3 tan (θ) + 1 = 0

Solve by substitution

2 tan^{2} (θ) - 3 tan (θ) + 1 = 0

Let:

tan (θ) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 3, c = 1

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

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

(- 3)^{2} - 4 \cdot 2 \cdot 1 = 1

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

\frac{- ( - 3 ) + 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 + 1}{2 \cdot 2}

Add the numbers:

3 + 1 = 4

= \frac{4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

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

\frac{- ( - 3 ) - 1}{2 \cdot 2}

Apply rule

- (- a) = a

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

Subtract the numbers:

3 - 1 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

u = 1, u = \frac{1}{2}

Substitute back

u = tan (θ)

tan (θ) = 1, tan (θ) = \frac{1}{2}

tan (θ) = 1, tan (θ) = \frac{1}{2}

tan (θ) = 1 : θ = \frac{π}{4} + πn

tan (θ) = 1

General solutions for

tan (θ) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

tan (θ) = \frac{1}{2} : θ = arctan (\frac{1}{2}) + πn

tan (θ) = \frac{1}{2}

Apply trig inverse properties

tan (θ) = \frac{1}{2}

General solutions for

tan (θ) = \frac{1}{2}

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

θ = arctan (\frac{1}{2}) + πn

θ = arctan (\frac{1}{2}) + πn

Combine all the solutions

θ = \frac{π}{4} + πn, θ = arctan (\frac{1}{2}) + πn

Show solutions in decimal form

θ = \frac{π}{4} + πn, θ = 0.46364 \dots + πn

Popular Examples

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cos (x) = \frac{9}{17}

-6sin(c)+0=sin(c)-3

- 6 sin (c) + 0 = sin (c) - 3

tan(x)= 11/12

tan (x) = \frac{11}{12}

2cos(a)+3=4

2 cos (a) + 3 = 4

tan(x)cot(x)-sqrt(3)tan(x)=0

tan (x) cot (x) - 3 tan (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 2tan^2(θ)-3tan(θ)+1=0 ?
The general solution for 2tan^2(θ)-3tan(θ)+1=0 is θ= pi/4+pin,θ=0.46364…+pin