What is the general solution for sin^2(θ)=6(cos(-θ)+1) ?

The general solution for sin^2(θ)=6(cos(-θ)+1) is θ=pi+2pin

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sin^2(θ)=6(cos(-θ)+1)

Solution

sin^{2} (θ) = 6 (cos (- θ) + 1)

Solution

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (θ) = 6 (cos (- θ) + 1)

Rewrite using trig identities

sin^{2} (θ) = 6 (cos (- θ) + 1)

sin^{2} (θ) = 6 (1 + cos (θ))

sin^{2} (θ) = 6 (1 + cos (θ))

Subtract

6 (1 + cos (θ))

from both sides

sin^{2} (θ) - 6 (1 + cos (θ)) = 0

Rewrite using trig identities

sin^{2} (θ) - (1 + cos (θ)) \cdot 6

Use the Pythagorean identity:

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

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

= 1 - cos^{2} (θ) - (1 + cos (θ)) \cdot 6

Simplify

1 - cos^{2} (θ) - (1 + cos (θ)) \cdot 6 : - cos^{2} (θ) - 6 cos (θ) - 5

1 - cos^{2} (θ) - (1 + cos (θ)) \cdot 6

= 1 - cos^{2} (θ) - 6 (1 + cos (θ))

Expand

- 6 (1 + cos (θ)) : - 6 - 6 cos (θ)

- 6 (1 + cos (θ))

Apply the distributive law:

a (b + c) = ab + a c

a = - 6, b = 1, c = cos (θ)

= - 6 \cdot 1 + (- 6) cos (θ)

Apply minus-plus rules

+ (- a) = - a

= - 6 \cdot 1 - 6 cos (θ)

Multiply the numbers:

6 \cdot 1 = 6

= - 6 - 6 cos (θ)

= 1 - cos^{2} (θ) - 6 - 6 cos (θ)

Simplify

1 - cos^{2} (θ) - 6 - 6 cos (θ) : - cos^{2} (θ) - 6 cos (θ) - 5

1 - cos^{2} (θ) - 6 - 6 cos (θ)

Group like terms

= - cos^{2} (θ) - 6 cos (θ) + 1 - 6

Add/Subtract the numbers:

1 - 6 = - 5

= - cos^{2} (θ) - 6 cos (θ) - 5

= - cos^{2} (θ) - 6 cos (θ) - 5

= - cos^{2} (θ) - 6 cos (θ) - 5

- 5 - cos^{2} (θ) - 6 cos (θ) = 0

Solve by substitution

- 5 - cos^{2} (θ) - 6 cos (θ) = 0

Let:

cos (θ) = u

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

- 5 - u^{2} - 6 u = 0 : u = - 5, u = - 1

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

Write in the standard form

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

- u^{2} - 6 u - 5 = 0

Solve with the quadratic formula

- u^{2} - 6 u - 5 = 0

Quadratic Equation Formula:

For

a = - 1, b = - 6, c = - 5

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 1 ) ( - 5 )}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 1 ) ( - 5 )}{2 ( - 1 )}

(- 6)^{2} - 4 (- 1) (- 5) = 4

(- 6)^{2} - 4 (- 1) (- 5)

Apply rule

- (- a) = a

= (- 6)^{2} - 4 \cdot 1 \cdot 5

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 1 \cdot 5

Multiply the numbers:

4 \cdot 1 \cdot 5 = 20

= 6^{2} - 20

6^{2} = 36

= 36 - 20

Subtract the numbers:

36 - 20 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

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

u = \frac{- ( - 6 ) + 4}{2 ( - 1 )} : - 5

\frac{- ( - 6 ) + 4}{2 ( - 1 )}

Remove parentheses:

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

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

Add the numbers:

6 + 4 = 10

= \frac{10}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{10}{- 2}

Apply the fraction rule:

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

= - \frac{10}{2}

Divide the numbers:

\frac{10}{2} = 5

= - 5

u = \frac{- ( - 6 ) - 4}{2 ( - 1 )} : - 1

\frac{- ( - 6 ) - 4}{2 ( - 1 )}

Remove parentheses:

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

= \frac{6 - 4}{- 2 \cdot 1}

Subtract the numbers:

6 - 4 = 2

= \frac{2}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{- 2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = - 5, u = - 1

Substitute back

u = cos (θ)

cos (θ) = - 5, cos (θ) = - 1

cos (θ) = - 5, cos (θ) = - 1

cos (θ) = - 5 :

No Solution

cos (θ) = - 5

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

General solutions for

cos (θ) = - 1

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 = π + 2 πn

θ = π + 2 πn

Combine all the solutions

θ = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(θ)=6(cos(-θ)+1) ?
The general solution for sin^2(θ)=6(cos(-θ)+1) is θ=pi+2pin