What is the general solution for -4cos^2(θ)-11cos(θ)-1=-7cos(θ) ?

The general solution for -4cos^2(θ)-11cos(θ)-1=-7cos(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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Popular >

-4cos^2(θ)-11cos(θ)-1=-7cos(θ)

Solution

- 4 cos^{2} (θ) - 11 cos (θ) - 1 = - 7 cos (θ)

Solution

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

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

- 4 cos^{2} (θ) - 11 cos (θ) - 1 = - 7 cos (θ)

Solve by substitution

- 4 cos^{2} (θ) - 11 cos (θ) - 1 = - 7 cos (θ)

Let:

cos (θ) = u

- 4 u^{2} - 11 u - 1 = - 7 u

- 4 u^{2} - 11 u - 1 = - 7 u : u = - \frac{1}{2}

- 4 u^{2} - 11 u - 1 = - 7 u

Move

7 u

to the left side

- 4 u^{2} - 11 u - 1 = - 7 u

Add

7 u

to both sides

- 4 u^{2} - 11 u - 1 + 7 u = - 7 u + 7 u

Simplify

- 4 u^{2} - 4 u - 1 = 0

- 4 u^{2} - 4 u - 1 = 0

Solve with the quadratic formula

- 4 u^{2} - 4 u - 1 = 0

Quadratic Equation Formula:

For

a = - 4, b = - 4, c = - 1

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

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

(- 4)^{2} - 4 (- 4) (- 1) = 0

(- 4)^{2} - 4 (- 4) (- 1)

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 ( - 4 )}

u = \frac{- ( - 4 )}{2 ( - 4 )}

\frac{- ( - 4 )}{2 ( - 4 )} = - \frac{1}{2}

\frac{- ( - 4 )}{2 ( - 4 )}

Remove parentheses:

(- a) = - a, - (- a) = a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{- 8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

u = - \frac{1}{2}

The solution to the quadratic equation is:

u = - \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Combine all the solutions

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

Popular Examples

0=1+cos(x/2)

0 = 1 + cos (\frac{x}{2})

sin(x)=cos(19)

sin (x) = cos (1 9^{\circ})

1/(1.5)=(sin(30))/(sin(x)),x<90

\frac{1}{1.5} = \frac{sin ( 3 0 ^{\circ} )}{sin ( x )}, x < 9 0^{\circ}

tan(A)= 8/15

tan (A) = \frac{8}{15}

tan^2(θ)=sec^2(θ)+1

tan^{2} (θ) = sec^{2} (θ) + 1

Frequently Asked Questions (FAQ)

What is the general solution for -4cos^2(θ)-11cos(θ)-1=-7cos(θ) ?
The general solution for -4cos^2(θ)-11cos(θ)-1=-7cos(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin