What is the general solution for 3-3sin(θ)=6cos^2(θ) ?

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

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3-3sin(θ)=6cos^2(θ)

Solution

3 - 3 sin (θ) = 6 cos^{2} (θ)

Solution

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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

3 - 3 sin (θ) = 6 cos^{2} (θ)

Subtract

6 cos^{2} (θ)

from both sides

3 - 3 sin (θ) - 6 cos^{2} (θ) = 0

Rewrite using trig identities

3 - 3 sin (θ) - 6 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= 3 - 3 sin (θ) - 6 (1 - sin^{2} (θ))

Simplify

3 - 3 sin (θ) - 6 (1 - sin^{2} (θ)) : 6 sin^{2} (θ) - 3 sin (θ) - 3

3 - 3 sin (θ) - 6 (1 - sin^{2} (θ))

Expand

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

- 6 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 6, b = 1, c = sin^{2} (θ)

= - 6 \cdot 1 - (- 6) sin^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 6 \cdot 1 + 6 sin^{2} (θ)

Multiply the numbers:

6 \cdot 1 = 6

= - 6 + 6 sin^{2} (θ)

= 3 - 3 sin (θ) - 6 + 6 sin^{2} (θ)

Simplify

3 - 3 sin (θ) - 6 + 6 sin^{2} (θ) : 6 sin^{2} (θ) - 3 sin (θ) - 3

3 - 3 sin (θ) - 6 + 6 sin^{2} (θ)

Group like terms

= - 3 sin (θ) + 6 sin^{2} (θ) + 3 - 6

Add/Subtract the numbers:

3 - 6 = - 3

= 6 sin^{2} (θ) - 3 sin (θ) - 3

= 6 sin^{2} (θ) - 3 sin (θ) - 3

= 6 sin^{2} (θ) - 3 sin (θ) - 3

- 3 - 3 sin (θ) + 6 sin^{2} (θ) = 0

Solve by substitution

- 3 - 3 sin (θ) + 6 sin^{2} (θ) = 0

Let:

sin (θ) = u

- 3 - 3 u + 6 u^{2} = 0

- 3 - 3 u + 6 u^{2} = 0 : u = 1, u = - \frac{1}{2}

- 3 - 3 u + 6 u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 6, b = - 3, c = - 3

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

(- 3)^{2} - 4 \cdot 6 (- 3) = 9

(- 3)^{2} - 4 \cdot 6 (- 3)

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 6 \cdot 3

Apply exponent rule:

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

n

is even

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

= 3^{2} + 4 \cdot 6 \cdot 3

Multiply the numbers:

4 \cdot 6 \cdot 3 = 72

= 3^{2} + 72

3^{2} = 9

= 9 + 72

Add the numbers:

9 + 72 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 3 ) \pm 9}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 9}{2 \cdot 6}, u_{2} = \frac{- ( - 3 ) - 9}{2 \cdot 6}

u = \frac{- ( - 3 ) + 9}{2 \cdot 6} : 1

\frac{- ( - 3 ) + 9}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{3 + 9}{2 \cdot 6}

Add the numbers:

3 + 9 = 12

= \frac{12}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{12}{12}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 3 ) - 9}{2 \cdot 6} : - \frac{1}{2}

\frac{- ( - 3 ) - 9}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{3 - 9}{2 \cdot 6}

Subtract the numbers:

3 - 9 = - 6

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{12}

Apply the fraction rule:

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

= - \frac{6}{12}

Cancel the common factor:

6

= - \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

sin (θ) = 1, sin (θ) = - \frac{1}{2}

sin (θ) = 1, sin (θ) = - \frac{1}{2}

sin (θ) = 1 : θ = \frac{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{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

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

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

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