What is the general solution for 2cos^2(θ)-9cos(θ)+4=0 ?

The general solution for 2cos^2(θ)-9cos(θ)+4=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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2cos^2(θ)-9cos(θ)+4=0

Solution

2 cos^{2} (θ) - 9 cos (θ) + 4 = 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) - 9 cos (θ) + 4 = 0

Solve by substitution

2 cos^{2} (θ) - 9 cos (θ) + 4 = 0

Let:

cos (θ) = u

2 u^{2} - 9 u + 4 = 0

2 u^{2} - 9 u + 4 = 0 : u = 4, u = \frac{1}{2}

2 u^{2} - 9 u + 4 = 0

Solve with the quadratic formula

2 u^{2} - 9 u + 4 = 0

Quadratic Equation Formula:

For

a = 2, b = - 9, c = 4

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

(- 9)^{2} - 4 \cdot 2 \cdot 4 = 7

(- 9)^{2} - 4 \cdot 2 \cdot 4

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

= 9^{2} - 4 \cdot 2 \cdot 4

Multiply the numbers:

4 \cdot 2 \cdot 4 = 32

= 9^{2} - 32

9^{2} = 81

= 81 - 32

Subtract the numbers:

81 - 32 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 9 ) \pm 7}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 9 ) + 7}{2 \cdot 2}, u_{2} = \frac{- ( - 9 ) - 7}{2 \cdot 2}

u = \frac{- ( - 9 ) + 7}{2 \cdot 2} : 4

\frac{- ( - 9 ) + 7}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{9 + 7}{2 \cdot 2}

Add the numbers:

9 + 7 = 16

= \frac{16}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{16}{4}

Divide the numbers:

\frac{16}{4} = 4

= 4

u = \frac{- ( - 9 ) - 7}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 9 ) - 7}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{9 - 7}{2 \cdot 2}

Subtract the numbers:

9 - 7 = 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 = 4, u = \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = 4, cos (θ) = \frac{1}{2}

cos (θ) = 4, cos (θ) = \frac{1}{2}

cos (θ) = 4 :

No Solution

cos (θ) = 4

- 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

Popular Examples

cos(2x+pi/4)=(sqrt(2))/2

cos (2 x + \frac{π}{4}) = \frac{2}{2}

solvefor x,tan(x)=1 solve for

x, tan (x) = 1

cot(t)=-sqrt(3)

cot (t) = - 3

tan(x)= 3/8

tan (x) = \frac{3}{8}

1-sin(x)=2cos^2(x)

1 - sin (x) = 2 cos^{2} (x)

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(θ)-9cos(θ)+4=0 ?
The general solution for 2cos^2(θ)-9cos(θ)+4=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin