What is the general solution for 4sin^2(θ/2)+8cos(θ/2)-7=0 ?

The general solution for 4sin^2(θ/2)+8cos(θ/2)-7=0 is θ=(2pi)/3+4pin,θ=(10pi)/3+4pin

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4sin^2(θ/2)+8cos(θ/2)-7=0

Solution

4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0

Solution

θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

Degrees

θ = 12 0^{\circ} + 72 0^{\circ} n, θ = 60 0^{\circ} + 72 0^{\circ} n

Solution steps

4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0

Rewrite using trig identities

- 7 + 4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2})

Use the Pythagorean identity:

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

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

= - 7 + 4 (1 - cos^{2} (\frac{θ}{2})) + 8 cos (\frac{θ}{2})

Simplify

- 7 + 4 (1 - cos^{2} (\frac{θ}{2})) + 8 cos (\frac{θ}{2}) : 8 cos (\frac{θ}{2}) - 4 cos^{2} (\frac{θ}{2}) - 3

- 7 + 4 (1 - cos^{2} (\frac{θ}{2})) + 8 cos (\frac{θ}{2})

Expand

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

4 (1 - cos^{2} (\frac{θ}{2}))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (\frac{θ}{2})

= 4 \cdot 1 - 4 cos^{2} (\frac{θ}{2})

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (\frac{θ}{2})

= - 7 + 4 - 4 cos^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2})

Add/Subtract the numbers:

- 7 + 4 = - 3

= 8 cos (\frac{θ}{2}) - 4 cos^{2} (\frac{θ}{2}) - 3

= 8 cos (\frac{θ}{2}) - 4 cos^{2} (\frac{θ}{2}) - 3

- 3 - 4 cos^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) = 0

Solve by substitution

- 3 - 4 cos^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) = 0

Let:

cos (\frac{θ}{2}) = u

- 3 - 4 u^{2} + 8 u = 0

- 3 - 4 u^{2} + 8 u = 0 : u = \frac{1}{2}, u = \frac{3}{2}

- 3 - 4 u^{2} + 8 u = 0

Write in the standard form

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

- 4 u^{2} + 8 u - 3 = 0

Solve with the quadratic formula

- 4 u^{2} + 8 u - 3 = 0

Quadratic Equation Formula:

For

a = - 4, b = 8, c = - 3

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) ( - 3 )}{2 ( - 4 )}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) ( - 3 )}{2 ( - 4 )}

8^{2} - 4 (- 4) (- 3) = 4

8^{2} - 4 (- 4) (- 3)

Apply rule

- (- a) = a

= 8^{2} - 4 \cdot 4 \cdot 3

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 8^{2} - 48

8^{2} = 64

= 64 - 48

Subtract the numbers:

64 - 48 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 8 + 4 = - 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}

u = \frac{- 8 - 4}{2 ( - 4 )} : \frac{3}{2}

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 8 - 4 = - 12

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 12}{- 8}

Apply the fraction rule:

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

= \frac{12}{8}

Cancel the common factor:

4

= \frac{3}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (\frac{θ}{2})

cos (\frac{θ}{2}) = \frac{1}{2}, cos (\frac{θ}{2}) = \frac{3}{2}

cos (\frac{θ}{2}) = \frac{1}{2}, cos (\frac{θ}{2}) = \frac{3}{2}

cos (\frac{θ}{2}) = \frac{1}{2} : θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

cos (\frac{θ}{2}) = \frac{1}{2}

General solutions for

cos (\frac{θ}{2}) = \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}

\frac{θ}{2} = \frac{π}{3} + 2 πn, \frac{θ}{2} = \frac{5 π}{3} + 2 πn

\frac{θ}{2} = \frac{π}{3} + 2 πn, \frac{θ}{2} = \frac{5 π}{3} + 2 πn

Solve

\frac{θ}{2} = \frac{π}{3} + 2 πn : θ = \frac{2 π}{3} + 4 πn

\frac{θ}{2} = \frac{π}{3} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{π}{3} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{π}{3} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{π}{3} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{π}{3} + 2 \cdot 2 πn : \frac{2 π}{3} + 4 πn

2 \cdot \frac{π}{3} + 2 \cdot 2 πn

Multiply

2 \cdot \frac{π}{3} : \frac{2 π}{3}

2 \cdot \frac{π}{3}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{3}

= \frac{2 π}{3} + 2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn

Solve

\frac{θ}{2} = \frac{5 π}{3} + 2 πn : θ = \frac{10 π}{3} + 4 πn

\frac{θ}{2} = \frac{5 π}{3} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{5 π}{3} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn : \frac{10 π}{3} + 4 πn

2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{3} = \frac{10 π}{3}

2 \cdot \frac{5 π}{3}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{5 π 2}{3}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

cos (\frac{θ}{2}) = \frac{3}{2} :

No Solution

cos (\frac{θ}{2}) = \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(θ/2)+8cos(θ/2)-7=0 ?
The general solution for 4sin^2(θ/2)+8cos(θ/2)-7=0 is θ=(2pi)/3+4pin,θ=(10pi)/3+4pin