What is the general solution for 2cos^2(θ)-3cos(θ)-2=0 ?

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

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2cos^2(θ)-3cos(θ)-2=0

Solution

2 cos^{2} (θ) - 3 cos (θ) - 2 = 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) - 3 cos (θ) - 2 = 0

Solve by substitution

2 cos^{2} (θ) - 3 cos (θ) - 2 = 0

Let:

cos (θ) = u

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

2 u^{2} - 3 u - 2 = 0 : u = 2, u = - \frac{1}{2}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 3, c = - 2

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

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

(- 3)^{2} - 4 \cdot 2 (- 2) = 5

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 2}, u_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 2}

u = \frac{- ( - 3 ) + 5}{2 \cdot 2} : 2

\frac{- ( - 3 ) + 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 + 5}{2 \cdot 2}

Add the numbers:

3 + 5 = 8

= \frac{8}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= 2

u = \frac{- ( - 3 ) - 5}{2 \cdot 2} : - \frac{1}{2}

\frac{- ( - 3 ) - 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 - 5}{2 \cdot 2}

Subtract the numbers:

3 - 5 = - 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 = 2, u = - \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = 2, cos (θ) = - \frac{1}{2}

cos (θ) = 2, cos (θ) = - \frac{1}{2}

cos (θ) = 2 :

No Solution

cos (θ) = 2

- 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

-7sin^2(θ)+4sin(θ)=-3

- 7 sin^{2} (θ) + 4 sin (θ) = - 3

sqrt(3)=sqrt(3)cos(2x)+sin(2x)

3 = 3 cos (2 x) + sin (2 x)

csc(θ)=-(2sqrt(3))/3

csc (θ) = - \frac{2 3}{3}

2sin(x/3)+sqrt(3)=0

2 sin (\frac{x}{3}) + 3 = 0

tan(θ)= 15/8

tan (θ) = \frac{15}{8}

Frequently Asked Questions (FAQ)

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