What is the general solution for 2cos^2(θ)-7cos(θ)-4=0 ?

The general solution for 2cos^2(θ)-7cos(θ)-4=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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2cos^2(θ)-7cos(θ)-4=0

Solution

2 cos^{2} (θ) - 7 cos (θ) - 4 = 0

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

2 cos^{2} (θ) - 7 cos (θ) - 4 = 0

Solve by substitution

2 cos^{2} (θ) - 7 cos (θ) - 4 = 0

Let:

cos (θ) = u

2 u^{2} - 7 u - 4 = 0

2 u^{2} - 7 u - 4 = 0 : u = 4, u = - \frac{1}{2}

2 u^{2} - 7 u - 4 = 0

Solve with the quadratic formula

2 u^{2} - 7 u - 4 = 0

Quadratic Equation Formula:

For

a = 2, b = - 7, c = - 4

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} + 4 \cdot 2 \cdot 4

Multiply the numbers:

4 \cdot 2 \cdot 4 = 32

= 7^{2} + 32

7^{2} = 49

= 49 + 32

Add the numbers:

49 + 32 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

7 - 9 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

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

= - \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{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

sin(θ)= 3/10 ,sin(θ+pi/6)

sin (θ) = \frac{3}{10}, sin (θ + \frac{π}{6})

3sin^2(x)=8cos(x)

3 sin^{2} (x) = 8 cos (x)

7cos(x)-24sin(x)=10

7 cos (x) - 24 sin (x) = 10

4cos(2θ)+1=2cos(θ)

4 cos (2 θ) + 1 = 2 cos (θ)

sin(2x)=-2cos^2(x),0<= x<= 2pi

sin (2 x) = - 2 cos^{2} (x), 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(θ)-7cos(θ)-4=0 ?
The general solution for 2cos^2(θ)-7cos(θ)-4=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin