What is the general solution for 3sin(θ)=1+cos(θ) ?

The general solution for 3sin(θ)=1+cos(θ) is θ=pi+2pin,θ=0.64350…+2pin

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Popular Trigonometry >

3sin(θ)=1+cos(θ)

Solution

3 sin (θ) = 1 + cos (θ)

Solution

θ = π + 2 πn, θ = 0.64350 \dots + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 36.86989 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (θ) = 1 + cos (θ)

Square both sides

(3 sin (θ))^{2} = (1 + cos (θ))^{2}

Subtract

(1 + cos (θ))^{2}

from both sides

9 sin^{2} (θ) - 1 - 2 cos (θ) - cos^{2} (θ) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

- 1 - cos^{2} (θ) - 2 cos (θ) + 9 (1 - cos^{2} (θ)) : - 10 cos^{2} (θ) - 2 cos (θ) + 8

- 1 - cos^{2} (θ) - 2 cos (θ) + 9 (1 - cos^{2} (θ))

Expand

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

9 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 9, b = 1, c = cos^{2} (θ)

= 9 \cdot 1 - 9 cos^{2} (θ)

Multiply the numbers:

9 \cdot 1 = 9

= 9 - 9 cos^{2} (θ)

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

Simplify

- 1 - cos^{2} (θ) - 2 cos (θ) + 9 - 9 cos^{2} (θ) : - 10 cos^{2} (θ) - 2 cos (θ) + 8

- 1 - cos^{2} (θ) - 2 cos (θ) + 9 - 9 cos^{2} (θ)

Group like terms

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

Add similar elements:

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

= - 10 cos^{2} (θ) - 2 cos (θ) - 1 + 9

Add/Subtract the numbers:

- 1 + 9 = 8

= - 10 cos^{2} (θ) - 2 cos (θ) + 8

= - 10 cos^{2} (θ) - 2 cos (θ) + 8

= - 10 cos^{2} (θ) - 2 cos (θ) + 8

8 - 10 cos^{2} (θ) - 2 cos (θ) = 0

Solve by substitution

8 - 10 cos^{2} (θ) - 2 cos (θ) = 0

Let:

cos (θ) = u

8 - 10 u^{2} - 2 u = 0

8 - 10 u^{2} - 2 u = 0 : u = - 1, u = \frac{4}{5}

8 - 10 u^{2} - 2 u = 0

Write in the standard form

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

- 10 u^{2} - 2 u + 8 = 0

Solve with the quadratic formula

- 10 u^{2} - 2 u + 8 = 0

Quadratic Equation Formula:

For

a = - 10, b = - 2, c = 8

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 10 ) \cdot 8}{2 ( - 10 )}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 10 ) \cdot 8}{2 ( - 10 )}

(- 2)^{2} - 4 (- 10) \cdot 8 = 18

(- 2)^{2} - 4 (- 10) \cdot 8

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 10 \cdot 8

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 10 \cdot 8

Multiply the numbers:

4 \cdot 10 \cdot 8 = 320

= 2^{2} + 320

2^{2} = 4

= 4 + 320

Add the numbers:

4 + 320 = 324

= 324

Factor the number:

324 = 1 8^{2}

= 1 8^{2}

Apply radical rule:

1 8^{2} = 18

= 18

u_{1, 2} = \frac{- ( - 2 ) \pm 18}{2 ( - 10 )}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 18}{2 ( - 10 )}, u_{2} = \frac{- ( - 2 ) - 18}{2 ( - 10 )}

u = \frac{- ( - 2 ) + 18}{2 ( - 10 )} : - 1

\frac{- ( - 2 ) + 18}{2 ( - 10 )}

Remove parentheses:

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

= \frac{2 + 18}{- 2 \cdot 10}

Add the numbers:

2 + 18 = 20

= \frac{20}{- 2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{20}{- 20}

Apply the fraction rule:

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

= - \frac{20}{20}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 2 ) - 18}{2 ( - 10 )} : \frac{4}{5}

\frac{- ( - 2 ) - 18}{2 ( - 10 )}

Remove parentheses:

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

= \frac{2 - 18}{- 2 \cdot 10}

Subtract the numbers:

2 - 18 = - 16

= \frac{- 16}{- 2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 16}{- 20}

Apply the fraction rule:

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

= \frac{16}{20}

Cancel the common factor:

4

= \frac{4}{5}

The solutions to the quadratic equation are:

u = - 1, u = \frac{4}{5}

Substitute back

u = cos (θ)

cos (θ) = - 1, cos (θ) = \frac{4}{5}

cos (θ) = - 1, cos (θ) = \frac{4}{5}

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

General solutions for

cos (θ) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = π + 2 πn

θ = π + 2 πn

cos (θ) = \frac{4}{5} : θ = arccos (\frac{4}{5}) + 2 πn, θ = 2 π - arccos (\frac{4}{5}) + 2 πn

cos (θ) = \frac{4}{5}

Apply trig inverse properties

cos (θ) = \frac{4}{5}

General solutions for

cos (θ) = \frac{4}{5}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{4}{5}) + 2 πn, θ = 2 π - arccos (\frac{4}{5}) + 2 πn

θ = arccos (\frac{4}{5}) + 2 πn, θ = 2 π - arccos (\frac{4}{5}) + 2 πn

Combine all the solutions

θ = π + 2 πn, θ = arccos (\frac{4}{5}) + 2 πn, θ = 2 π - arccos (\frac{4}{5}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

3 sin (θ) = 1 + cos (θ)

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

Check the solution

π + 2 πn :

True

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

3 sin (θ) = 1 + cos (θ)

plug in

θ = π + 2 π 1

3 sin (π + 2 π 1) = 1 + cos (π + 2 π 1)

Refine

0 = 0

\Rightarrow True

Check the solution

arccos (\frac{4}{5}) + 2 πn :

True

arccos (\frac{4}{5}) + 2 πn

Plug in

n = 1

arccos (\frac{4}{5}) + 2 π 1

For

3 sin (θ) = 1 + cos (θ)

plug in

θ = arccos (\frac{4}{5}) + 2 π 1

3 sin (arccos (\frac{4}{5}) + 2 π 1) = 1 + cos (arccos (\frac{4}{5}) + 2 π 1)

Refine

1.8 = 1.8

\Rightarrow True

Check the solution

2 π - arccos (\frac{4}{5}) + 2 πn :

False

2 π - arccos (\frac{4}{5}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{4}{5}) + 2 π 1

For

3 sin (θ) = 1 + cos (θ)

plug in

θ = 2 π - arccos (\frac{4}{5}) + 2 π 1

3 sin (2 π - arccos (\frac{4}{5}) + 2 π 1) = 1 + cos (2 π - arccos (\frac{4}{5}) + 2 π 1)

Refine

- 1.8 = 1.8

\Rightarrow False

θ = π + 2 πn, θ = arccos (\frac{4}{5}) + 2 πn

Show solutions in decimal form

θ = π + 2 πn, θ = 0.64350 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3sin(θ)=1+cos(θ) ?
The general solution for 3sin(θ)=1+cos(θ) is θ=pi+2pin,θ=0.64350…+2pin