What is the general solution for -3cos(2θ)-18sin(θ)-8=-5sin(θ)-6 ?

The general solution for -3cos(2θ)-18sin(θ)-8=-5sin(θ)-6 is θ=-0.33983…+2pin,θ=pi+0.33983…+2pin

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Popular >

-3cos(2θ)-18sin(θ)-8=-5sin(θ)-6

Solution

- 3 cos (2 θ) - 18 sin (θ) - 8 = - 5 sin (θ) - 6

Solution

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

Degrees

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

Solution steps

- 3 cos (2 θ) - 18 sin (θ) - 8 = - 5 sin (θ) - 6

Subtract

- 5 sin (θ) - 6

from both sides

- 3 cos (2 θ) - 13 sin (θ) - 2 = 0

Rewrite using trig identities

- 2 - 13 sin (θ) - 3 cos (2 θ)

Use the Double Angle identity:

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

= - 2 - 13 sin (θ) - 3 (1 - 2 sin^{2} (θ))

Simplify

- 2 - 13 sin (θ) - 3 (1 - 2 sin^{2} (θ)) : 6 sin^{2} (θ) - 13 sin (θ) - 5

- 2 - 13 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 - 13 sin (θ) - 3 + 6 sin^{2} (θ)

Simplify

- 2 - 13 sin (θ) - 3 + 6 sin^{2} (θ) : 6 sin^{2} (θ) - 13 sin (θ) - 5

- 2 - 13 sin (θ) - 3 + 6 sin^{2} (θ)

Group like terms

= - 13 sin (θ) + 6 sin^{2} (θ) - 2 - 3

Subtract the numbers:

- 2 - 3 = - 5

= 6 sin^{2} (θ) - 13 sin (θ) - 5

= 6 sin^{2} (θ) - 13 sin (θ) - 5

= 6 sin^{2} (θ) - 13 sin (θ) - 5

- 5 - 13 sin (θ) + 6 sin^{2} (θ) = 0

Solve by substitution

- 5 - 13 sin (θ) + 6 sin^{2} (θ) = 0

Let:

sin (θ) = u

- 5 - 13 u + 6 u^{2} = 0

- 5 - 13 u + 6 u^{2} = 0 : u = \frac{5}{2}, u = - \frac{1}{3}

- 5 - 13 u + 6 u^{2} = 0

Write in the standard form

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

6 u^{2} - 13 u - 5 = 0

Solve with the quadratic formula

6 u^{2} - 13 u - 5 = 0

Quadratic Equation Formula:

For

a = 6, b = - 13, c = - 5

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 6 ( - 5 )}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 6 ( - 5 )}{2 \cdot 6}

(- 13)^{2} - 4 \cdot 6 (- 5) = 17

(- 13)^{2} - 4 \cdot 6 (- 5)

Apply rule

- (- a) = a

= (- 13)^{2} + 4 \cdot 6 \cdot 5

Apply exponent rule:

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

n

is even

(- 13)^{2} = 1 3^{2}

= 1 3^{2} + 4 \cdot 6 \cdot 5

Multiply the numbers:

4 \cdot 6 \cdot 5 = 120

= 1 3^{2} + 120

1 3^{2} = 169

= 169 + 120

Add the numbers:

169 + 120 = 289

= 289

Factor the number:

289 = 1 7^{2}

= 1 7^{2}

Apply radical rule:

1 7^{2} = 17

= 17

u_{1, 2} = \frac{- ( - 13 ) \pm 17}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- ( - 13 ) + 17}{2 \cdot 6}, u_{2} = \frac{- ( - 13 ) - 17}{2 \cdot 6}

u = \frac{- ( - 13 ) + 17}{2 \cdot 6} : \frac{5}{2}

\frac{- ( - 13 ) + 17}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{13 + 17}{2 \cdot 6}

Add the numbers:

13 + 17 = 30

= \frac{30}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{30}{12}

Cancel the common factor:

6

= \frac{5}{2}

u = \frac{- ( - 13 ) - 17}{2 \cdot 6} : - \frac{1}{3}

\frac{- ( - 13 ) - 17}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{13 - 17}{2 \cdot 6}

Subtract the numbers:

13 - 17 = - 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{5}{2}, u = - \frac{1}{3}

Substitute back

u = sin (θ)

sin (θ) = \frac{5}{2}, sin (θ) = - \frac{1}{3}

sin (θ) = \frac{5}{2}, sin (θ) = - \frac{1}{3}

sin (θ) = \frac{5}{2} :

No Solution

sin (θ) = \frac{5}{2}

- 1 \leq sin (x) \leq 1

No Solution

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

Show solutions in decimal form

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

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Frequently Asked Questions (FAQ)

What is the general solution for -3cos(2θ)-18sin(θ)-8=-5sin(θ)-6 ?
The general solution for -3cos(2θ)-18sin(θ)-8=-5sin(θ)-6 is θ=-0.33983…+2pin,θ=pi+0.33983…+2pin