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

The general solution for 2tan^2(θ)-tan(θ)=3 is θ=0.98279…+pin,θ=(3pi)/4+pin

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2tan^2(θ)-tan(θ)=3

Solution

2 tan^{2} (θ) - tan (θ) = 3

Solution

θ = 0.98279 \dots + πn, θ = \frac{3 π}{4} + πn

Degrees

θ = 56.30993 \dots^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

2 tan^{2} (θ) - tan (θ) = 3

Solve by substitution

2 tan^{2} (θ) - tan (θ) = 3

Let:

tan (θ) = u

2 u^{2} - u = 3

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

2 u^{2} - u = 3

Move

3

to the left side

2 u^{2} - u = 3

Subtract

3

from both sides

2 u^{2} - u - 3 = 3 - 3

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 3 = 24

4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

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

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

\frac{- ( - 1 ) + 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 5}{2 \cdot 2}

Add the numbers:

1 + 5 = 6

= \frac{6}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{6}{4}

Cancel the common factor:

2

= \frac{3}{2}

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

\frac{- ( - 1 ) - 5}{2 \cdot 2}

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 5 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{3}{2}, u = - 1

Substitute back

u = tan (θ)

tan (θ) = \frac{3}{2}, tan (θ) = - 1

tan (θ) = \frac{3}{2}, tan (θ) = - 1

tan (θ) = \frac{3}{2} : θ = arctan (\frac{3}{2}) + πn

tan (θ) = \frac{3}{2}

Apply trig inverse properties

tan (θ) = \frac{3}{2}

General solutions for

tan (θ) = \frac{3}{2}

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

θ = arctan (\frac{3}{2}) + πn

θ = arctan (\frac{3}{2}) + πn

tan (θ) = - 1 : θ = \frac{3 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{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Combine all the solutions

θ = arctan (\frac{3}{2}) + πn, θ = \frac{3 π}{4} + πn

Show solutions in decimal form

θ = 0.98279 \dots + πn, θ = \frac{3 π}{4} + πn

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170 = \frac{9 0 ^{2}}{16} sin (x) cos (x)

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sin (x - \frac{π}{3}) = 0

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\frac{- 2 + 3}{2} = - 1 + cos (4 θ + 135)

Frequently Asked Questions (FAQ)

What is the general solution for 2tan^2(θ)-tan(θ)=3 ?
The general solution for 2tan^2(θ)-tan(θ)=3 is θ=0.98279…+pin,θ=(3pi)/4+pin