What is the general solution for 2cos^2(θ)-11cos(θ)+5=0 ?

The general solution for 2cos^2(θ)-11cos(θ)+5=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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2cos^2(θ)-11cos(θ)+5=0

Solution

2 cos^{2} (θ) - 11 cos (θ) + 5 = 0

Solution

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (θ) - 11 cos (θ) + 5 = 0

Solve by substitution

2 cos^{2} (θ) - 11 cos (θ) + 5 = 0

Let:

cos (θ) = u

2 u^{2} - 11 u + 5 = 0

2 u^{2} - 11 u + 5 = 0 : u = 5, u = \frac{1}{2}

2 u^{2} - 11 u + 5 = 0

Solve with the quadratic formula

2 u^{2} - 11 u + 5 = 0

Quadratic Equation Formula:

For

a = 2, b = - 11, c = 5

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

(- 11)^{2} - 4 \cdot 2 \cdot 5 = 9

(- 11)^{2} - 4 \cdot 2 \cdot 5

Apply exponent rule:

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

n

is even

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

= 1 1^{2} - 4 \cdot 2 \cdot 5

Multiply the numbers:

4 \cdot 2 \cdot 5 = 40

= 1 1^{2} - 40

1 1^{2} = 121

= 121 - 40

Subtract the numbers:

121 - 40 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 11 ) \pm 9}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 11 ) + 9}{2 \cdot 2}, u_{2} = \frac{- ( - 11 ) - 9}{2 \cdot 2}

u = \frac{- ( - 11 ) + 9}{2 \cdot 2} : 5

\frac{- ( - 11 ) + 9}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{11 + 9}{2 \cdot 2}

Add the numbers:

11 + 9 = 20

= \frac{20}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{20}{4}

Divide the numbers:

\frac{20}{4} = 5

= 5

u = \frac{- ( - 11 ) - 9}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 11 ) - 9}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{11 - 9}{2 \cdot 2}

Subtract the numbers:

11 - 9 = 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 = 5, u = \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = 5, cos (θ) = \frac{1}{2}

cos (θ) = 5, cos (θ) = \frac{1}{2}

cos (θ) = 5 :

No Solution

cos (θ) = 5

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 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{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Combine all the solutions

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(θ)-11cos(θ)+5=0 ?
The general solution for 2cos^2(θ)-11cos(θ)+5=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin