What is the general solution for sin(θ)=1-2cos(θ) ?

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

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sin(θ)=1-2cos(θ)

Solution

sin (θ) = 1 - 2 cos (θ)

Solution

θ = \frac{π}{2} + 2 πn, θ = 2 π - 0.64350 \dots + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 323.13010 \dots^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) = 1 - 2 cos (θ)

Square both sides

sin^{2} (θ) = (1 - 2 cos (θ))^{2}

Subtract

(1 - 2 cos (θ))^{2}

from both sides

sin^{2} (θ) - 1 + 4 cos (θ) - 4 cos^{2} (θ) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= 4 cos (θ) - 4 cos^{2} (θ) - cos^{2} (θ)

Simplify

= 4 cos (θ) - 5 cos^{2} (θ)

4 cos (θ) - 5 cos^{2} (θ) = 0

Solve by substitution

4 cos (θ) - 5 cos^{2} (θ) = 0

Let:

cos (θ) = u

4 u - 5 u^{2} = 0

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

4 u - 5 u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 5, b = 4, c = 0

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 5 ) \cdot 0}{2 ( - 5 )}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 5 ) \cdot 0}{2 ( - 5 )}

4^{2} - 4 (- 5) \cdot 0 = 4

4^{2} - 4 (- 5) \cdot 0

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 5 \cdot 0

Apply rule

0 \cdot a = 0

= 4^{2} + 0

4^{2} + 0 = 4^{2}

= 4^{2}

Apply radical rule:

assuming

a \geq 0

= 4

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

Separate the solutions

u_{1} = \frac{- 4 + 4}{2 ( - 5 )}, u_{2} = \frac{- 4 - 4}{2 ( - 5 )}

u = \frac{- 4 + 4}{2 ( - 5 )} : 0

\frac{- 4 + 4}{2 ( - 5 )}

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 4 + 4 = 0

= \frac{0}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{0}{- 10}

Apply the fraction rule:

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

= - \frac{0}{10}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

\frac{- 4 - 4}{2 ( - 5 )}

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 4 - 4 = - 8

= \frac{- 8}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 8}{- 10}

Apply the fraction rule:

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

= \frac{8}{10}

Cancel the common factor:

2

= \frac{4}{5}

The solutions to the quadratic equation are:

u = 0, u = \frac{4}{5}

Substitute back

u = cos (θ)

cos (θ) = 0, cos (θ) = \frac{4}{5}

cos (θ) = 0, cos (θ) = \frac{4}{5}

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

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 = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 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 = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 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

sin (θ) = 1 - 2 cos (θ)

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

Check the solution

\frac{π}{2} + 2 πn :

True

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

sin (θ) = 1 - 2 cos (θ)

plug in

θ = \frac{π}{2} + 2 π 1

sin (\frac{π}{2} + 2 π 1) = 1 - 2 cos (\frac{π}{2} + 2 π 1)

Refine

1 = 1

\Rightarrow True

Check the solution

\frac{3 π}{2} + 2 πn :

False

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

sin (θ) = 1 - 2 cos (θ)

plug in

θ = \frac{3 π}{2} + 2 π 1

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

Refine

- 1 = 1

\Rightarrow False

Check the solution

arccos (\frac{4}{5}) + 2 πn :

False

arccos (\frac{4}{5}) + 2 πn

Plug in

n = 1

arccos (\frac{4}{5}) + 2 π 1

For

sin (θ) = 1 - 2 cos (θ)

plug in

θ = arccos (\frac{4}{5}) + 2 π 1

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

Refine

0.6 = - 0.6

\Rightarrow False

Check the solution

2 π - arccos (\frac{4}{5}) + 2 πn :

True

2 π - arccos (\frac{4}{5}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{4}{5}) + 2 π 1

For

sin (θ) = 1 - 2 cos (θ)

plug in

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

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

Refine

- 0.6 = - 0.6

\Rightarrow True

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

Show solutions in decimal form

θ = \frac{π}{2} + 2 πn, θ = 2 π - 0.64350 \dots + 2 πn

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Frequently Asked Questions (FAQ)

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