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Popular >

6cos^2(θ)-5=6sin^2(θ)+sin(θ)

Solution

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

Solution

θ = - 0.33983 \dots + 2 πn, θ = π + 0.33983 \dots + 2 πn, θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

Degrees

θ = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 199.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 14.47751 \dots^{\circ} + 36 0^{\circ} n, θ = 165.52248 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

6 sin^{2} (θ) + sin (θ)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= - 5 - sin (θ) + 6 (1 - sin^{2} (θ)) - 6 sin^{2} (θ)

Simplify

- 5 - sin (θ) + 6 (1 - sin^{2} (θ)) - 6 sin^{2} (θ) : - 12 sin^{2} (θ) - sin (θ) + 1

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

Multiply the numbers:

6 \cdot 1 = 6

= 6 - 6 sin^{2} (θ)

= - 5 - sin (θ) + 6 - 6 sin^{2} (θ) - 6 sin^{2} (θ)

Simplify

- 5 - sin (θ) + 6 - 6 sin^{2} (θ) - 6 sin^{2} (θ) : - 12 sin^{2} (θ) - sin (θ) + 1

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

Add similar elements:

- 6 sin^{2} (θ) - 6 sin^{2} (θ) = - 12 sin^{2} (θ)

= - 5 - sin (θ) + 6 - 12 sin^{2} (θ)

Group like terms

= - sin (θ) - 12 sin^{2} (θ) - 5 + 6

Add/Subtract the numbers:

- 5 + 6 = 1

= - 12 sin^{2} (θ) - sin (θ) + 1

= - 12 sin^{2} (θ) - sin (θ) + 1

= - 12 sin^{2} (θ) - sin (θ) + 1

1 - sin (θ) - 12 sin^{2} (θ) = 0

Solve by substitution

1 - sin (θ) - 12 sin^{2} (θ) = 0

Let:

sin (θ) = u

1 - u - 12 u^{2} = 0

1 - u - 12 u^{2} = 0 : u = - \frac{1}{3}, u = \frac{1}{4}

1 - u - 12 u^{2} = 0

Write in the standard form

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

- 12 u^{2} - u + 1 = 0

Solve with the quadratic formula

- 12 u^{2} - u + 1 = 0

Quadratic Equation Formula:

For

a = - 12, b = - 1, c = 1

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 12 ) \cdot 1}{2 ( - 12 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 12 ) \cdot 1}{2 ( - 12 )}

(- 1)^{2} - 4 (- 12) \cdot 1 = 7

(- 1)^{2} - 4 (- 12) \cdot 1

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 12 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 12 \cdot 1 = 48

4 \cdot 12 \cdot 1

Multiply the numbers:

4 \cdot 12 \cdot 1 = 48

= 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 ( - 12 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 7}{2 ( - 12 )}, u_{2} = \frac{- ( - 1 ) - 7}{2 ( - 12 )}

u = \frac{- ( - 1 ) + 7}{2 ( - 12 )} : - \frac{1}{3}

\frac{- ( - 1 ) + 7}{2 ( - 12 )}

Remove parentheses:

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

= \frac{1 + 7}{- 2 \cdot 12}

Add the numbers:

1 + 7 = 8

= \frac{8}{- 2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{8}{- 24}

Apply the fraction rule:

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

= - \frac{8}{24}

Cancel the common factor:

8

= - \frac{1}{3}

u = \frac{- ( - 1 ) - 7}{2 ( - 12 )} : \frac{1}{4}

\frac{- ( - 1 ) - 7}{2 ( - 12 )}

Remove parentheses:

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

= \frac{1 - 7}{- 2 \cdot 12}

Subtract the numbers:

1 - 7 = - 6

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

Multiply the numbers:

2 \cdot 12 = 24

= \frac{- 6}{- 24}

Apply the fraction rule:

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

= \frac{6}{24}

Cancel the common factor:

6

= \frac{1}{4}

The solutions to the quadratic equation are:

u = - \frac{1}{3}, u = \frac{1}{4}

Substitute back

u = sin (θ)

sin (θ) = - \frac{1}{3}, sin (θ) = \frac{1}{4}

sin (θ) = - \frac{1}{3}, sin (θ) = \frac{1}{4}

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

sin (θ) = - \frac{1}{3}

Apply trig inverse properties

sin (θ) = - \frac{1}{3}

General solutions for

sin (θ) = - \frac{1}{3}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn

θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn

sin (θ) = \frac{1}{4} : θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

sin (θ) = \frac{1}{4}

Apply trig inverse properties

sin (θ) = \frac{1}{4}

General solutions for

sin (θ) = \frac{1}{4}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

Combine all the solutions

θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn, θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

Show solutions in decimal form

θ = - 0.33983 \dots + 2 πn, θ = π + 0.33983 \dots + 2 πn, θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

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