What is the general solution for 6sin^2(θ/2)=cos^2(θ)+5 ?

The general solution for 6sin^2(θ/2)=cos^2(θ)+5 is θ=pi+2pin

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Popular >

6sin^2(θ/2)=cos^2(θ)+5

Solution

6 sin^{2} (\frac{θ}{2}) = cos^{2} (θ) + 5

Solution

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

6 sin^{2} (\frac{θ}{2}) = cos^{2} (θ) + 5

Subtract

cos^{2} (θ) + 5

from both sides

6 sin^{2} (\frac{θ}{2}) - cos^{2} (θ) - 5 = 0

Rewrite using trig identities

- 5 - cos^{2} (θ) + 6 sin^{2} (\frac{θ}{2})

Use the following identity:

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

Use the Double Angle identity

cos (2 θ) = 1 - 2 sin^{2} (θ)

Switch sides

2 sin^{2} (θ) - 1 = - cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 - cos (2 θ)

Divide both sides by

2

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

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

Simplify

- 5 - cos^{2} (θ) + 6 \cdot \frac{1 - cos ( 2 \cdot \frac{θ}{2} )}{2} : - cos^{2} (θ) - 3 cos (θ) - 2

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

6 \cdot \frac{1 - cos ( 2 \cdot \frac{θ}{2} )}{2} = 3 (- cos (θ) + 1)

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

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

\frac{1 - cos ( 2 \cdot \frac{θ}{2} )}{2}

Multiply

2 \cdot \frac{θ}{2} : θ

2 \cdot \frac{θ}{2}

Multiply fractions:

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

= \frac{θ \cdot 2}{2}

Cancel the common factor:

2

= θ

= \frac{1 - cos ( θ )}{2}

= 6 \cdot \frac{- cos ( θ ) + 1}{2}

Multiply fractions:

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

= \frac{( 1 - cos ( θ ) ) \cdot 6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3 (- cos (θ) + 1)

= - 5 - cos^{2} (θ) + 3 (- cos (θ) + 1)

Expand

3 (- cos (θ) + 1) : - 3 cos (θ) + 3

3 (- cos (θ) + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = - cos (θ), c = 1

= 3 (- cos (θ)) + 3 \cdot 1

Apply minus-plus rules

+ (- a) = - a

= - 3 cos (θ) + 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= - 3 cos (θ) + 3

= - 5 - cos^{2} (θ) - 3 cos (θ) + 3

Simplify

- 5 - cos^{2} (θ) - 3 cos (θ) + 3 : - cos^{2} (θ) - 3 cos (θ) - 2

- 5 - cos^{2} (θ) - 3 cos (θ) + 3

Group like terms

= - cos^{2} (θ) - 3 cos (θ) - 5 + 3

Add/Subtract the numbers:

- 5 + 3 = - 2

= - cos^{2} (θ) - 3 cos (θ) - 2

= - cos^{2} (θ) - 3 cos (θ) - 2

= - cos^{2} (θ) - 3 cos (θ) - 2

- 2 - cos^{2} (θ) - 3 cos (θ) = 0

Solve by substitution

- 2 - cos^{2} (θ) - 3 cos (θ) = 0

Let:

cos (θ) = u

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

- 2 - u^{2} - 3 u = 0 : u = - 2, u = - 1

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 1, b = - 3, c = - 2

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

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

(- 3)^{2} - 4 (- 1) (- 2) = 1

(- 3)^{2} - 4 (- 1) (- 2)

Apply rule

- (- a) = a

= (- 3)^{2} - 4 \cdot 1 \cdot 2

Apply exponent rule:

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

n

is even

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

= 3^{2} - 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

\frac{- ( - 3 ) + 1}{2 ( - 1 )}

Remove parentheses:

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

= \frac{3 + 1}{- 2 \cdot 1}

Add the numbers:

3 + 1 = 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{- 2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

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

\frac{- ( - 3 ) - 1}{2 ( - 1 )}

Remove parentheses:

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

= \frac{3 - 1}{- 2 \cdot 1}

Subtract the numbers:

3 - 1 = 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 = - 2, u = - 1

Substitute back

u = cos (θ)

cos (θ) = - 2, cos (θ) = - 1

cos (θ) = - 2, cos (θ) = - 1

cos (θ) = - 2 :

No Solution

cos (θ) = - 2

- 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 6sin^2(θ/2)=cos^2(θ)+5 ?
The general solution for 6sin^2(θ/2)=cos^2(θ)+5 is θ=pi+2pin