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

The general solution for 3*cos(θ)=2-sin(θ) is θ=-0.56432…+2pin,θ=1.20782…+2pin

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Popular Trigonometry >

3*cos(θ)=2-sin(θ)

Solution

3 \cdot cos (θ) = 2 - sin (θ)

Solution

θ = - 0.56432 \dots + 2 πn, θ = 1.20782 \dots + 2 πn

Degrees

θ = - 32.33353 \dots^{\circ} + 36 0^{\circ} n, θ = 69.20342 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 cos (θ) = 2 - sin (θ)

Square both sides

(3 cos (θ))^{2} = (2 - sin (θ))^{2}

Subtract

(2 - sin (θ))^{2}

from both sides

9 cos^{2} (θ) - 4 + 4 sin (θ) - sin^{2} (θ) = 0

Rewrite using trig identities

- 4 - sin^{2} (θ) + 4 sin (θ) + 9 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= - 4 - sin^{2} (θ) + 4 sin (θ) + 9 (1 - sin^{2} (θ))

Simplify

- 4 - sin^{2} (θ) + 4 sin (θ) + 9 (1 - sin^{2} (θ)) : 4 sin (θ) - 10 sin^{2} (θ) + 5

- 4 - sin^{2} (θ) + 4 sin (θ) + 9 (1 - sin^{2} (θ))

Expand

9 (1 - sin^{2} (θ)) : 9 - 9 sin^{2} (θ)

9 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 9, b = 1, c = sin^{2} (θ)

= 9 \cdot 1 - 9 sin^{2} (θ)

Multiply the numbers:

9 \cdot 1 = 9

= 9 - 9 sin^{2} (θ)

= - 4 - sin^{2} (θ) + 4 sin (θ) + 9 - 9 sin^{2} (θ)

Simplify

- 4 - sin^{2} (θ) + 4 sin (θ) + 9 - 9 sin^{2} (θ) : 4 sin (θ) - 10 sin^{2} (θ) + 5

- 4 - sin^{2} (θ) + 4 sin (θ) + 9 - 9 sin^{2} (θ)

Group like terms

= - sin^{2} (θ) + 4 sin (θ) - 9 sin^{2} (θ) - 4 + 9

Add similar elements:

- sin^{2} (θ) - 9 sin^{2} (θ) = - 10 sin^{2} (θ)

= - 10 sin^{2} (θ) + 4 sin (θ) - 4 + 9

Add/Subtract the numbers:

- 4 + 9 = 5

= 4 sin (θ) - 10 sin^{2} (θ) + 5

= 4 sin (θ) - 10 sin^{2} (θ) + 5

= 4 sin (θ) - 10 sin^{2} (θ) + 5

5 - 10 sin^{2} (θ) + 4 sin (θ) = 0

Solve by substitution

5 - 10 sin^{2} (θ) + 4 sin (θ) = 0

Let:

sin (θ) = u

5 - 10 u^{2} + 4 u = 0

5 - 10 u^{2} + 4 u = 0 : u = - \frac{- 2 + 3 6}{10}, u = \frac{2 + 3 6}{10}

5 - 10 u^{2} + 4 u = 0

Write in the standard form

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

- 10 u^{2} + 4 u + 5 = 0

Solve with the quadratic formula

- 10 u^{2} + 4 u + 5 = 0

Quadratic Equation Formula:

For

a = - 10, b = 4, c = 5

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 10 ) \cdot 5}{2 ( - 10 )}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 10 ) \cdot 5}{2 ( - 10 )}

4^{2} - 4 (- 10) \cdot 5 = 66

4^{2} - 4 (- 10) \cdot 5

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 10 \cdot 5

Multiply the numbers:

4 \cdot 10 \cdot 5 = 200

= 4^{2} + 200

4^{2} = 16

= 16 + 200

Add the numbers:

16 + 200 = 216

= 216

Prime factorization of

216 : 2^{3} \cdot 3^{3}

216

216

divides by

2216 = 108 \cdot 2

= 2 \cdot 108

108

divides by

2108 = 54 \cdot 2

= 2 \cdot 2 \cdot 54

54

divides by

254 = 27 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 27

27

divides by

327 = 9 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 9

9

divides by

39 = 3 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3

= 2^{3} \cdot 3^{3}

= 2^{3} \cdot 3^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 3^{2} \cdot 2 \cdot 3

Apply radical rule:

n ab = n a n b

= 2^{2} 3^{2} 2 \cdot 3

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2 3^{2} 2 \cdot 3

Apply radical rule:

n a^{n} = a

3^{2} = 3

= 2 \cdot 3 2 \cdot 3

Refine

= 66

u_{1, 2} = \frac{- 4 \pm 6 6}{2 ( - 10 )}

Separate the solutions

u_{1} = \frac{- 4 + 6 6}{2 ( - 10 )}, u_{2} = \frac{- 4 - 6 6}{2 ( - 10 )}

u = \frac{- 4 + 6 6}{2 ( - 10 )} : - \frac{- 2 + 3 6}{10}

\frac{- 4 + 6 6}{2 ( - 10 )}

Remove parentheses:

(- a) = - a

= \frac{- 4 + 6 6}{- 2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 4 + 6 6}{- 20}

Apply the fraction rule:

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

= - \frac{- 4 + 6 6}{20}

Cancel

\frac{- 4 + 6 6}{20} : \frac{3 6 - 2}{10}

\frac{- 4 + 6 6}{20}

Factor

- 4 + 66 : 2 (- 2 + 36)

- 4 + 66

Rewrite as

= - 2 \cdot 2 + 2 \cdot 36

Factor out common term

2

= 2 (- 2 + 36)

= \frac{2 ( - 2 + 3 6 )}{20}

Cancel the common factor:

2

= \frac{- 2 + 3 6}{10}

= - \frac{3 6 - 2}{10}

= - \frac{- 2 + 3 6}{10}

u = \frac{- 4 - 6 6}{2 ( - 10 )} : \frac{2 + 3 6}{10}

\frac{- 4 - 6 6}{2 ( - 10 )}

Remove parentheses:

(- a) = - a

= \frac{- 4 - 6 6}{- 2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 4 - 6 6}{- 20}

Apply the fraction rule:

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

- 4 - 66 = - (4 + 66)

= \frac{4 + 6 6}{20}

Factor

4 + 66 : 2 (2 + 36)

4 + 66

Rewrite as

= 2 \cdot 2 + 2 \cdot 36

Factor out common term

2

= 2 (2 + 36)

= \frac{2 ( 2 + 3 6 )}{20}

Cancel the common factor:

2

= \frac{2 + 3 6}{10}

The solutions to the quadratic equation are:

u = - \frac{- 2 + 3 6}{10}, u = \frac{2 + 3 6}{10}

Substitute back

u = sin (θ)

sin (θ) = - \frac{- 2 + 3 6}{10}, sin (θ) = \frac{2 + 3 6}{10}

sin (θ) = - \frac{- 2 + 3 6}{10}, sin (θ) = \frac{2 + 3 6}{10}

sin (θ) = - \frac{- 2 + 3 6}{10} : θ = arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn, θ = π + arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

sin (θ) = - \frac{- 2 + 3 6}{10}

Apply trig inverse properties

sin (θ) = - \frac{- 2 + 3 6}{10}

General solutions for

sin (θ) = - \frac{- 2 + 3 6}{10}

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

θ = arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn, θ = π + arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

θ = arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn, θ = π + arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

sin (θ) = \frac{2 + 3 6}{10} : θ = arcsin (\frac{2 + 3 6}{10}) + 2 πn, θ = π - arcsin (\frac{2 + 3 6}{10}) + 2 πn

sin (θ) = \frac{2 + 3 6}{10}

Apply trig inverse properties

sin (θ) = \frac{2 + 3 6}{10}

General solutions for

sin (θ) = \frac{2 + 3 6}{10}

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

θ = arcsin (\frac{2 + 3 6}{10}) + 2 πn, θ = π - arcsin (\frac{2 + 3 6}{10}) + 2 πn

θ = arcsin (\frac{2 + 3 6}{10}) + 2 πn, θ = π - arcsin (\frac{2 + 3 6}{10}) + 2 πn

Combine all the solutions

θ = arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn, θ = π + arcsin (\frac{- 2 + 3 6}{10}) + 2 πn, θ = arcsin (\frac{2 + 3 6}{10}) + 2 πn, θ = π - arcsin (\frac{2 + 3 6}{10}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

3 cos (θ) = 2 - sin (θ)

Remove the ones that don't agree with the equation.

Check the solution

arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn :

True

arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn

Plug in

n = 1

arcsin (- \frac{- 2 + 3 6}{10}) + 2 π 1

For

3 cos (θ) = 2 - sin (θ)

plug in

θ = arcsin (- \frac{- 2 + 3 6}{10}) + 2 π 1

3 cos (arcsin (- \frac{- 2 + 3 6}{10}) + 2 π 1) = 2 - sin (arcsin (- \frac{- 2 + 3 6}{10}) + 2 π 1)

Refine

2.53484 \dots = 2.53484 \dots

\Rightarrow True

Check the solution

π + arcsin (\frac{- 2 + 3 6}{10}) + 2 πn :

False

π + arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

Plug in

n = 1

π + arcsin (\frac{- 2 + 3 6}{10}) + 2 π 1

For

3 cos (θ) = 2 - sin (θ)

plug in

θ = π + arcsin (\frac{- 2 + 3 6}{10}) + 2 π 1

3 cos (π + arcsin (\frac{- 2 + 3 6}{10}) + 2 π 1) = 2 - sin (π + arcsin (\frac{- 2 + 3 6}{10}) + 2 π 1)

Refine

- 2.53484 \dots = 2.53484 \dots

\Rightarrow False

Check the solution

arcsin (\frac{2 + 3 6}{10}) + 2 πn :

True

arcsin (\frac{2 + 3 6}{10}) + 2 πn

Plug in

n = 1

arcsin (\frac{2 + 3 6}{10}) + 2 π 1

For

3 cos (θ) = 2 - sin (θ)

plug in

θ = arcsin (\frac{2 + 3 6}{10}) + 2 π 1

3 cos (arcsin (\frac{2 + 3 6}{10}) + 2 π 1) = 2 - sin (arcsin (\frac{2 + 3 6}{10}) + 2 π 1)

Refine

1.06515 \dots = 1.06515 \dots

\Rightarrow True

Check the solution

π - arcsin (\frac{2 + 3 6}{10}) + 2 πn :

False

π - arcsin (\frac{2 + 3 6}{10}) + 2 πn

Plug in

n = 1

π - arcsin (\frac{2 + 3 6}{10}) + 2 π 1

For

3 cos (θ) = 2 - sin (θ)

plug in

θ = π - arcsin (\frac{2 + 3 6}{10}) + 2 π 1

3 cos (π - arcsin (\frac{2 + 3 6}{10}) + 2 π 1) = 2 - sin (π - arcsin (\frac{2 + 3 6}{10}) + 2 π 1)

Refine

- 1.06515 \dots = 1.06515 \dots

\Rightarrow False

θ = arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn, θ = arcsin (\frac{2 + 3 6}{10}) + 2 πn

Show solutions in decimal form

θ = - 0.56432 \dots + 2 πn, θ = 1.20782 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3*cos(θ)=2-sin(θ) ?
The general solution for 3*cos(θ)=2-sin(θ) is θ=-0.56432…+2pin,θ=1.20782…+2pin