What is the general solution for 6+6sin(θ)=4cos^2(θ) ?

The general solution for 6+6sin(θ)=4cos^2(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=(3pi)/2+2pin

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6+6sin(θ)=4cos^2(θ)

Solution

6 + 6 sin (θ) = 4 cos^{2} (θ)

Solution

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

6 + 6 sin (θ) = 4 cos^{2} (θ)

Subtract

4 cos^{2} (θ)

from both sides

6 + 6 sin (θ) - 4 cos^{2} (θ) = 0

Rewrite using trig identities

6 - 4 cos^{2} (θ) + 6 sin (θ)

Use the Pythagorean identity:

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

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

= 6 - 4 (1 - sin^{2} (θ)) + 6 sin (θ)

Simplify

6 - 4 (1 - sin^{2} (θ)) + 6 sin (θ) : 4 sin^{2} (θ) + 6 sin (θ) + 2

6 - 4 (1 - sin^{2} (θ)) + 6 sin (θ)

Expand

- 4 (1 - sin^{2} (θ)) : - 4 + 4 sin^{2} (θ)

- 4 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (θ)

= - 4 \cdot 1 - (- 4) sin^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 sin^{2} (θ)

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 sin^{2} (θ)

= 6 - 4 + 4 sin^{2} (θ) + 6 sin (θ)

Subtract the numbers:

6 - 4 = 2

= 4 sin^{2} (θ) + 6 sin (θ) + 2

= 4 sin^{2} (θ) + 6 sin (θ) + 2

2 + 4 sin^{2} (θ) + 6 sin (θ) = 0

Solve by substitution

2 + 4 sin^{2} (θ) + 6 sin (θ) = 0

Let:

sin (θ) = u

2 + 4 u^{2} + 6 u = 0

2 + 4 u^{2} + 6 u = 0 : u = - \frac{1}{2}, u = - 1

2 + 4 u^{2} + 6 u = 0

Write in the standard form

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

4 u^{2} + 6 u + 2 = 0

Solve with the quadratic formula

4 u^{2} + 6 u + 2 = 0

Quadratic Equation Formula:

For

a = 4, b = 6, c = 2

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

6^{2} - 4 \cdot 4 \cdot 2 = 2

6^{2} - 4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 6^{2} - 32

6^{2} = 36

= 36 - 32

Subtract the numbers:

36 - 32 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- 6 \pm 2}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 6 + 2}{2 \cdot 4}, u_{2} = \frac{- 6 - 2}{2 \cdot 4}

u = \frac{- 6 + 2}{2 \cdot 4} : - \frac{1}{2}

\frac{- 6 + 2}{2 \cdot 4}

Add/Subtract the numbers:

- 6 + 2 = - 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{- 6 - 2}{2 \cdot 4} : - 1

\frac{- 6 - 2}{2 \cdot 4}

Subtract the numbers:

- 6 - 2 = - 8

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8}{8}

Apply the fraction rule:

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

= - \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

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

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

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

General solutions for

sin (θ) = - \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

sin (θ) = - 1

General solutions for

sin (θ) = - 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 6+6sin(θ)=4cos^2(θ) ?
The general solution for 6+6sin(θ)=4cos^2(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=(3pi)/2+2pin