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Popular >

-3cos(2θ)-sin(θ)-6=-8

Solution

- 3 cos (2 θ) - sin (θ) - 6 = - 8

Solution

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = - 0.33983 \dots + 2 πn, θ = π + 0.33983 \dots + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n, θ = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 199.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 3 cos (2 θ) - sin (θ) - 6 = - 8

Subtract

- 8

from both sides

- 3 cos (2 θ) - sin (θ) + 2 = 0

Rewrite using trig identities

2 - sin (θ) - 3 cos (2 θ)

Use the Double Angle identity:

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

= 2 - sin (θ) - 3 (1 - 2 sin^{2} (θ))

Simplify

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

2 - sin (θ) - 3 (1 - 2 sin^{2} (θ))

Expand

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

- 3 (1 - 2 sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 1, c = 2 sin^{2} (θ)

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

Apply minus-plus rules

- (- a) = a

= - 3 \cdot 1 + 3 \cdot 2 sin^{2} (θ)

Simplify

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

- 3 \cdot 1 + 3 \cdot 2 sin^{2} (θ)

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 \cdot 2 sin^{2} (θ)

Multiply the numbers:

3 \cdot 2 = 6

= - 3 + 6 sin^{2} (θ)

= - 3 + 6 sin^{2} (θ)

= 2 - sin (θ) - 3 + 6 sin^{2} (θ)

Simplify

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

2 - sin (θ) - 3 + 6 sin^{2} (θ)

Group like terms

= - sin (θ) + 6 sin^{2} (θ) + 2 - 3

Add/Subtract the numbers:

2 - 3 = - 1

= 6 sin^{2} (θ) - sin (θ) - 1

= 6 sin^{2} (θ) - sin (θ) - 1

= 6 sin^{2} (θ) - sin (θ) - 1

- 1 - sin (θ) + 6 sin^{2} (θ) = 0

Solve by substitution

- 1 - sin (θ) + 6 sin^{2} (θ) = 0

Let:

sin (θ) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 6 \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 6 \cdot 1 = 24

4 \cdot 6 \cdot 1

Multiply the numbers:

4 \cdot 6 \cdot 1 = 24

= 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 6}, u_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 6}

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

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

Apply rule

- (- a) = a

= \frac{1 + 5}{2 \cdot 6}

Add the numbers:

1 + 5 = 6

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{6}{12}

Cancel the common factor:

6

= \frac{1}{2}

u = \frac{- ( - 1 ) - 5}{2 \cdot 6} : - \frac{1}{3}

\frac{- ( - 1 ) - 5}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{1 - 5}{2 \cdot 6}

Subtract the numbers:

1 - 5 = - 4

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 4}{12}

Apply the fraction rule:

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

= - \frac{4}{12}

Cancel the common factor:

4

= - \frac{1}{3}

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = - \frac{1}{3}

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{3}

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{3}

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 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{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

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

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn

Show solutions in decimal form

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = - 0.33983 \dots + 2 πn, θ = π + 0.33983 \dots + 2 πn

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