What is the general solution for 4sin^2(θ)-8cos(θ)+1=0,0<= θ<2pi ?

Q: What is the general solution for 4sin^2(θ)-8cos(θ)+1=0,0<= θ<2pi ?

The general solution for 4sin^2(θ)-8cos(θ)+1=0,0<= θ<2pi is θ= pi/3 ,θ=(5pi)/3

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4sin^2(θ)-8cos(θ)+1=0,0<= θ<2pi

Solution

4 sin^{2} (θ) - 8 cos (θ) + 1 = 0, 0 \leq θ < 2 π

Solution

θ = \frac{π}{3}, θ = \frac{5 π}{3}

Degrees

θ = 6 0^{\circ}, θ = 30 0^{\circ}

Solution steps

4 sin^{2} (θ) - 8 cos (θ) + 1 = 0, 0 \leq θ < 2 π

Rewrite using trig identities

1 + 4 sin^{2} (θ) - 8 cos (θ)

Use the Pythagorean identity:

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

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

= 1 + 4 (1 - cos^{2} (θ)) - 8 cos (θ)

Simplify

1 + 4 (1 - cos^{2} (θ)) - 8 cos (θ) : - 4 cos^{2} (θ) - 8 cos (θ) + 5

1 + 4 (1 - cos^{2} (θ)) - 8 cos (θ)

Expand

4 (1 - cos^{2} (θ)) : 4 - 4 cos^{2} (θ)

4 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (θ)

= 4 \cdot 1 - 4 cos^{2} (θ)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (θ)

= 1 + 4 - 4 cos^{2} (θ) - 8 cos (θ)

Add the numbers:

1 + 4 = 5

= - 4 cos^{2} (θ) - 8 cos (θ) + 5

= - 4 cos^{2} (θ) - 8 cos (θ) + 5

5 - 4 cos^{2} (θ) - 8 cos (θ) = 0

Solve by substitution

5 - 4 cos^{2} (θ) - 8 cos (θ) = 0

Let:

cos (θ) = u

5 - 4 u^{2} - 8 u = 0

5 - 4 u^{2} - 8 u = 0 : u = - \frac{5}{2}, u = \frac{1}{2}

5 - 4 u^{2} - 8 u = 0

Write in the standard form

a x^{2} + b x + c = 0

- 4 u^{2} - 8 u + 5 = 0

Solve with the quadratic formula

- 4 u^{2} - 8 u + 5 = 0

Quadratic Equation Formula:

For

a = - 4, b = - 8, c = 5

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

(- 8)^{2} - 4 (- 4) \cdot 5 = 12

(- 8)^{2} - 4 (- 4) \cdot 5

Apply rule

- (- a) = a

= (- 8)^{2} + 4 \cdot 4 \cdot 5

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 4 \cdot 5

Multiply the numbers:

4 \cdot 4 \cdot 5 = 80

= 8^{2} + 80

8^{2} = 64

= 64 + 80

Add the numbers:

64 + 80 = 144

= 144

Factor the number:

144 = 1 2^{2}

= 1 2^{2}

Apply radical rule:

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- ( - 8 ) \pm 12}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 12}{2 ( - 4 )}, u_{2} = \frac{- ( - 8 ) - 12}{2 ( - 4 )}

u = \frac{- ( - 8 ) + 12}{2 ( - 4 )} : - \frac{5}{2}

\frac{- ( - 8 ) + 12}{2 ( - 4 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{8 + 12}{- 2 \cdot 4}

Add the numbers:

8 + 12 = 20

= \frac{20}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{20}{- 8}

Apply the fraction rule:

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

= - \frac{20}{8}

Cancel the common factor:

4

= - \frac{5}{2}

u = \frac{- ( - 8 ) - 12}{2 ( - 4 )} : \frac{1}{2}

\frac{- ( - 8 ) - 12}{2 ( - 4 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{8 - 12}{- 2 \cdot 4}

Subtract the numbers:

8 - 12 = - 4

= \frac{- 4}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = - \frac{5}{2}, cos (θ) = \frac{1}{2}

cos (θ) = - \frac{5}{2}, cos (θ) = \frac{1}{2}

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

No Solution

cos (θ) = - \frac{5}{2}, 0 \leq θ < 2 π

- 1 \leq cos (x) \leq 1

No Solution

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

cos (θ) = \frac{1}{2}, 0 \leq θ < 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 < 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 4sin^2(θ)-8cos(θ)+1=0,0<= θ<2pi ?
The general solution for 4sin^2(θ)-8cos(θ)+1=0,0<= θ<2pi is θ= pi/3 ,θ=(5pi)/3