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

Q: What is the general solution for 2cos^2(θ)-7cos(θ)+3=0,0<= θ<= 2pi ?

The general solution for 2cos^2(θ)-7cos(θ)+3=0,0<= θ<= 2pi is θ= pi/3 ,θ=(5pi)/3

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2cos^2(θ)-7cos(θ)+3=0,0<= θ<= 2pi

Solution

2 cos^{2} (θ) - 7 cos (θ) + 3 = 0, 0 \leq θ \leq 2 π

Solution

θ = \frac{π}{3}, θ = \frac{5 π}{3}

Degrees

θ = 6 0^{\circ}, θ = 30 0^{\circ}

Solution steps

2 cos^{2} (θ) - 7 cos (θ) + 3 = 0, 0 \leq θ \leq 2 π

Solve by substitution

2 cos^{2} (θ) - 7 cos (θ) + 3 = 0

Let:

cos (θ) = u

2 u^{2} - 7 u + 3 = 0

2 u^{2} - 7 u + 3 = 0 : u = 3, u = \frac{1}{2}

2 u^{2} - 7 u + 3 = 0

Solve with the quadratic formula

2 u^{2} - 7 u + 3 = 0

Quadratic Equation Formula:

For

a = 2, b = - 7, c = 3

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

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

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

Apply exponent rule:

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

n

is even

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

= 7^{2} - 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

Subtract the numbers:

49 - 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

7 - 5 = 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 = 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, 0 \leq θ \leq 2 π :

No Solution

cos (θ) = 3, 0 \leq θ \leq 2 π

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = \frac{1}{2}, 0 \leq θ \leq 2 π : θ = \frac{π}{3}, θ = \frac{5 π}{3}

cos (θ) = \frac{1}{2}, 0 \leq θ \leq 2 π

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

Combine all the solutions

θ = \frac{π}{3}, θ = \frac{5 π}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(θ)-7cos(θ)+3=0,0<= θ<= 2pi ?
The general solution for 2cos^2(θ)-7cos(θ)+3=0,0<= θ<= 2pi is θ= pi/3 ,θ=(5pi)/3