What is the general solution for 4tan^2(θ)=11tan(θ)+3 ?

The general solution for 4tan^2(θ)=11tan(θ)+3 is θ=1.24904…+pin,θ=-0.24497…+pin

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Popular >

4tan^2(θ)=11tan(θ)+3

Solution

4 tan^{2} (θ) = 11 tan (θ) + 3

Solution

θ = 1.24904 \dots + πn, θ = - 0.24497 \dots + πn

Degrees

θ = 71.56505 \dots^{\circ} + 18 0^{\circ} n, θ = - 14.03624 \dots^{\circ} + 18 0^{\circ} n

Solution steps

4 tan^{2} (θ) = 11 tan (θ) + 3

Solve by substitution

4 tan^{2} (θ) = 11 tan (θ) + 3

Let:

tan (θ) = u

4 u^{2} = 11 u + 3

4 u^{2} = 11 u + 3 : u = 3, u = - \frac{1}{4}

4 u^{2} = 11 u + 3

Move

3

to the left side

4 u^{2} = 11 u + 3

Subtract

3

from both sides

4 u^{2} - 3 = 11 u + 3 - 3

Simplify

4 u^{2} - 3 = 11 u

4 u^{2} - 3 = 11 u

Move

11 u

to the left side

4 u^{2} - 3 = 11 u

Subtract

11 u

from both sides

4 u^{2} - 3 - 11 u = 11 u - 11 u

Simplify

4 u^{2} - 3 - 11 u = 0

4 u^{2} - 3 - 11 u = 0

Write in the standard form

a x^{2} + b x + c = 0

4 u^{2} - 11 u - 3 = 0

Solve with the quadratic formula

4 u^{2} - 11 u - 3 = 0

Quadratic Equation Formula:

For

a = 4, b = - 11, c = - 3

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

(- 11)^{2} - 4 \cdot 4 (- 3) = 13

(- 11)^{2} - 4 \cdot 4 (- 3)

Apply rule

- (- a) = a

= (- 11)^{2} + 4 \cdot 4 \cdot 3

Apply exponent rule:

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

n

is even

(- 11)^{2} = 1 1^{2}

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

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 1 1^{2} + 48

1 1^{2} = 121

= 121 + 48

Add the numbers:

121 + 48 = 169

= 169

Factor the number:

169 = 1 3^{2}

= 1 3^{2}

Apply radical rule:

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- ( - 11 ) \pm 13}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 11 ) + 13}{2 \cdot 4}, u_{2} = \frac{- ( - 11 ) - 13}{2 \cdot 4}

u = \frac{- ( - 11 ) + 13}{2 \cdot 4} : 3

\frac{- ( - 11 ) + 13}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{11 + 13}{2 \cdot 4}

Add the numbers:

11 + 13 = 24

= \frac{24}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{24}{8}

Divide the numbers:

\frac{24}{8} = 3

= 3

u = \frac{- ( - 11 ) - 13}{2 \cdot 4} : - \frac{1}{4}

\frac{- ( - 11 ) - 13}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{11 - 13}{2 \cdot 4}

Subtract the numbers:

11 - 13 = - 2

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2}{8}

Apply the fraction rule:

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

= - \frac{2}{8}

Cancel the common factor:

2

= - \frac{1}{4}

The solutions to the quadratic equation are:

u = 3, u = - \frac{1}{4}

Substitute back

u = tan (θ)

tan (θ) = 3, tan (θ) = - \frac{1}{4}

tan (θ) = 3, tan (θ) = - \frac{1}{4}

tan (θ) = 3 : θ = arctan (3) + πn

tan (θ) = 3

Apply trig inverse properties

tan (θ) = 3

General solutions for

tan (θ) = 3

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

θ = arctan (3) + πn

θ = arctan (3) + πn

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

tan (θ) = - \frac{1}{4}

Apply trig inverse properties

tan (θ) = - \frac{1}{4}

General solutions for

tan (θ) = - \frac{1}{4}

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

θ = arctan (- \frac{1}{4}) + πn

θ = arctan (- \frac{1}{4}) + πn

Combine all the solutions

θ = arctan (3) + πn, θ = arctan (- \frac{1}{4}) + πn

Show solutions in decimal form

θ = 1.24904 \dots + πn, θ = - 0.24497 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4tan^2(θ)=11tan(θ)+3 ?
The general solution for 4tan^2(θ)=11tan(θ)+3 is θ=1.24904…+pin,θ=-0.24497…+pin