What is the general solution for 2cos^2(θ)-5cos(θ)-3=0 ?

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

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2cos^2(θ)-5cos(θ)-3=0

Solution

2 cos^{2} (θ) - 5 cos (θ) - 3 = 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) - 5 cos (θ) - 3 = 0

Solve by substitution

2 cos^{2} (θ) - 5 cos (θ) - 3 = 0

Let:

cos (θ) = u

2 u^{2} - 5 u - 3 = 0

2 u^{2} - 5 u - 3 = 0 : u = 3, u = - \frac{1}{2}

2 u^{2} - 5 u - 3 = 0

Solve with the quadratic formula

2 u^{2} - 5 u - 3 = 0

Quadratic Equation Formula:

For

a = 2, b = - 5, c = - 3

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 (- 3) = 7

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 5 ) \pm 7}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 7}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 7}{2 \cdot 2}

u = \frac{- ( - 5 ) + 7}{2 \cdot 2} : 3

\frac{- ( - 5 ) + 7}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 + 7}{2 \cdot 2}

Add the numbers:

5 + 7 = 12

= \frac{12}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= 3

u = \frac{- ( - 5 ) - 7}{2 \cdot 2} : - \frac{1}{2}

\frac{- ( - 5 ) - 7}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 - 7}{2 \cdot 2}

Subtract the numbers:

5 - 7 = - 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 = 3, u = - \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = 3, cos (θ) = - \frac{1}{2}

cos (θ) = 3, cos (θ) = - \frac{1}{2}

cos (θ) = 3 :

No Solution

cos (θ) = 3

- 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

15sin(x)+6cos(x)-3=0

15 sin (x) + 6 cos (x) - 3 = 0

sqrt(1+cot^2(x))=8

1 + cot^{2} (x) = 8

1/(1-sin(x))+1/(1+sin(x))=2csc^2(x)

\frac{1}{1 - sin ( x )} + \frac{1}{1 + sin ( x )} = 2 csc^{2} (x)

tan(x)-sqrt(1-2tan^2(x))=0

tan (x) - 1 - 2 tan^{2} (x) = 0

cos(θ)=-sqrt(2/5),tan(2θ)

cos (θ) = - \frac{2}{5}, tan (2 θ)

Frequently Asked Questions (FAQ)

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