What is the general solution for sin(θ)=-(sqrt(5))/3 cos(2θ) ?

The general solution for sin(θ)=-(sqrt(5))/3 cos(2θ) is θ=-0.46364…+2pin,θ=pi+0.46364…+2pin

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Popular Trigonometry >

sin(θ)=-(sqrt(5))/3 cos(2θ)

Solution

sin (θ) = - \frac{5}{3} cos (2 θ)

Solution

θ = - 0.46364 \dots + 2 πn, θ = π + 0.46364 \dots + 2 πn

Degrees

θ = - 26.56505 \dots^{\circ} + 36 0^{\circ} n, θ = 206.56505 \dots^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) = - \frac{5}{3} cos (2 θ)

Subtract

- \frac{5}{3} cos (2 θ)

from both sides

sin (θ) + \frac{5}{3} cos (2 θ) = 0

Simplify

sin (θ) + \frac{5}{3} cos (2 θ) : \frac{3 sin ( θ ) + 5 cos ( 2 θ )}{3}

sin (θ) + \frac{5}{3} cos (2 θ)

Multiply

\frac{5}{3} cos (2 θ) : \frac{5 cos ( 2 θ )}{3}

\frac{5}{3} cos (2 θ)

Multiply fractions:

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

= \frac{5 cos ( 2 θ )}{3}

= sin (θ) + \frac{5 cos ( 2 θ )}{3}

Convert element to fraction:

sin (θ) = \frac{sin ( θ ) 3}{3}

= \frac{sin ( θ ) \cdot 3}{3} + \frac{5 cos ( 2 θ )}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( θ ) \cdot 3 + 5 cos ( 2 θ )}{3}

\frac{3 sin ( θ ) + 5 cos ( 2 θ )}{3} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

3 sin (θ) + 5 cos (2 θ) = 0

Rewrite using trig identities

3 sin (θ) + cos (2 θ) 5

Use the Double Angle identity:

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

= 3 sin (θ) + 5 (1 - 2 sin^{2} (θ))

(1 - 2 sin^{2} (θ)) 5 + 3 sin (θ) = 0

Solve by substitution

(1 - 2 sin^{2} (θ)) 5 + 3 sin (θ) = 0

Let:

sin (θ) = u

(1 - 2 u^{2}) 5 + 3 u = 0

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

(1 - 2 u^{2}) 5 + 3 u = 0

Expand

(1 - 2 u^{2}) 5 + 3 u : 5 - 25 u^{2} + 3 u

(1 - 2 u^{2}) 5 + 3 u

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

Expand

5 (1 - 2 u^{2}) : 5 - 25 u^{2}

5 (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = 1, c = 2 u^{2}

= 5 \cdot 1 - 5 \cdot 2 u^{2}

= 1 \cdot 5 - 25 u^{2}

Multiply:

1 \cdot 5 = 5

= 5 - 25 u^{2}

= 5 - 25 u^{2} + 3 u

5 - 25 u^{2} + 3 u = 0

Write in the standard form

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

- 25 u^{2} + 3 u + 5 = 0

Solve with the quadratic formula

- 25 u^{2} + 3 u + 5 = 0

Quadratic Equation Formula:

For

a = - 25, b = 3, c = 5

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 5 ) 5}{2 ( - 2 5 )}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 5 ) 5}{2 ( - 2 5 )}

3^{2} - 4 (- 25) 5 = 7

3^{2} - 4 (- 25) 5

Apply rule

- (- a) = a

= 3^{2} + 4 \cdot 255

4 \cdot 255 = 40

4 \cdot 255

Multiply the numbers:

4 \cdot 2 = 8

= 855

Apply radical rule:

a a = a

55 = 5

= 8 \cdot 5

Multiply the numbers:

8 \cdot 5 = 40

= 40

= 3^{2} + 40

3^{2} = 9

= 9 + 40

Add the numbers:

9 + 40 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 3 \pm 7}{2 ( - 2 5 )}

Separate the solutions

u_{1} = \frac{- 3 + 7}{2 ( - 2 5 )}, u_{2} = \frac{- 3 - 7}{2 ( - 2 5 )}

u = \frac{- 3 + 7}{2 ( - 2 5 )} : - \frac{5}{5}

\frac{- 3 + 7}{2 ( - 2 5 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 + 7}{- 2 \cdot 2 5}

Add/Subtract the numbers:

- 3 + 7 = 4

= \frac{4}{- 2 \cdot 2 5}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{- 4 5}

Apply the fraction rule:

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

= - \frac{4}{4 5}

Divide the numbers:

\frac{4}{4} = 1

= - \frac{1}{5}

Rationalize

- \frac{1}{5} : - \frac{5}{5}

- \frac{1}{5}

Multiply by the conjugate

\frac{5}{5}

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

1 \cdot 5 = 5

55 = 5

55

Apply radical rule:

a a = a

55 = 5

= 5

= - \frac{5}{5}

= - \frac{5}{5}

u = \frac{- 3 - 7}{2 ( - 2 5 )} : \frac{5}{2}

\frac{- 3 - 7}{2 ( - 2 5 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 - 7}{- 2 \cdot 2 5}

Subtract the numbers:

- 3 - 7 = - 10

= \frac{- 10}{- 2 \cdot 2 5}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 10}{- 4 5}

Apply the fraction rule:

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

= \frac{10}{4 5}

Cancel the common factor:

2

= \frac{5}{2 5}

Apply radical rule:

5 = 5^{\frac{1}{2}}

= \frac{5}{2 \cdot 5 ^{\frac{1}{2}}}

Apply exponent rule:

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

\frac{5 ^{1}}{5 ^{\frac{1}{2}}} = 5^{1 - \frac{1}{2}}

= \frac{5 ^{1 - \frac{1}{2}}}{2}

Subtract the numbers:

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

= \frac{5 ^{\frac{1}{2}}}{2}

Apply radical rule:

5^{\frac{1}{2}} = 5

= \frac{5}{2}

The solutions to the quadratic equation are:

u = - \frac{5}{5}, u = \frac{5}{2}

Substitute back

u = sin (θ)

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

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

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

sin (θ) = - \frac{5}{5}

Apply trig inverse properties

sin (θ) = - \frac{5}{5}

General solutions for

sin (θ) = - \frac{5}{5}

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

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

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

sin (θ) = \frac{5}{2} :

No Solution

sin (θ) = \frac{5}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

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

Show solutions in decimal form

θ = - 0.46364 \dots + 2 πn, θ = π + 0.46364 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sin(θ)=-(sqrt(5))/3 cos(2θ) ?
The general solution for sin(θ)=-(sqrt(5))/3 cos(2θ) is θ=-0.46364…+2pin,θ=pi+0.46364…+2pin