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

The general solution for 1-2sin(θ)=cos(θ) is θ=pi-0.92729…+2pin,θ=2pin

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Popular >

1-2sin(θ)=cos(θ)

Solution

1 - 2 sin (θ) = cos (θ)

Solution

θ = π - 0.92729 \dots + 2 πn, θ = 2 πn

Degrees

θ = 126.86989 \dots^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

1 - 2 sin (θ) = cos (θ)

Square both sides

(1 - 2 sin (θ))^{2} = cos^{2} (θ)

Subtract

cos^{2} (θ)

from both sides

(1 - 2 sin (θ))^{2} - cos^{2} (θ) = 0

Rewrite using trig identities

(1 - 2 sin (θ))^{2} - cos^{2} (θ)

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = 2 sin (θ)

= 1^{2} - 2 \cdot 1 \cdot 2 sin (θ) + (2 sin (θ))^{2}

Simplify

1^{2} - 2 \cdot 1 \cdot 2 sin (θ) + (2 sin (θ))^{2} : 1 - 4 sin (θ) + 4 sin^{2} (θ)

1^{2} - 2 \cdot 1 \cdot 2 sin (θ) + (2 sin (θ))^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 2 \cdot 1 \cdot 2 sin (θ) + (2 sin (θ))^{2}

2 \cdot 1 \cdot 2 sin (θ) = 4 sin (θ)

2 \cdot 1 \cdot 2 sin (θ)

Multiply the numbers:

2 \cdot 1 \cdot 2 = 4

= 4 sin (θ)

(2 sin (θ))^{2} = 4 sin^{2} (θ)

(2 sin (θ))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} sin^{2} (θ)

2^{2} = 4

= 4 sin^{2} (θ)

= 1 - 4 sin (θ) + 4 sin^{2} (θ)

= 1 - 4 sin (θ) + 4 sin^{2} (θ)

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

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

- (1 - sin^{2} (θ))

Distribute parentheses

= - (1) - (- sin^{2} (θ))

Apply minus-plus rules

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

= - 1 + sin^{2} (θ)

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

Simplify

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

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

Group like terms

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

Add similar elements:

4 sin^{2} (θ) + sin^{2} (θ) = 5 sin^{2} (θ)

= - 4 sin (θ) + 5 sin^{2} (θ) + 1 - 1

1 - 1 = 0

= 5 sin^{2} (θ) - 4 sin (θ)

= 5 sin^{2} (θ) - 4 sin (θ)

= 5 sin^{2} (θ) - 4 sin (θ)

- 4 sin (θ) + 5 sin^{2} (θ) = 0

Solve by substitution

- 4 sin (θ) + 5 sin^{2} (θ) = 0

Let:

sin (θ) = u

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

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

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

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

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

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

Apply exponent rule:

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

n

is even

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

= 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 \cdot 5}

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

4 + 4 = 8

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{8}{10}

Cancel the common factor:

2

= \frac{4}{5}

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

4 - 4 = 0

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{0}{10}

Apply rule

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

= 0

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

sin (θ) = \frac{4}{5}, sin (θ) = 0

sin (θ) = \frac{4}{5}, sin (θ) = 0

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

sin (θ) = \frac{4}{5}

Apply trig inverse properties

sin (θ) = \frac{4}{5}

General solutions for

sin (θ) = \frac{4}{5}

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

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

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

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 0

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

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

1 - 2 sin (θ) = cos (θ)

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

Check the solution

arcsin (\frac{4}{5}) + 2 πn :

False

arcsin (\frac{4}{5}) + 2 πn

Plug in

n = 1

arcsin (\frac{4}{5}) + 2 π 1

For

1 - 2 sin (θ) = cos (θ)

plug in

θ = arcsin (\frac{4}{5}) + 2 π 1

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

Refine

- 0.6 = 0.6

\Rightarrow False

Check the solution

π - arcsin (\frac{4}{5}) + 2 πn :

True

π - arcsin (\frac{4}{5}) + 2 πn

Plug in

n = 1

π - arcsin (\frac{4}{5}) + 2 π 1

For

1 - 2 sin (θ) = cos (θ)

plug in

θ = π - arcsin (\frac{4}{5}) + 2 π 1

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

Refine

- 0.6 = - 0.6

\Rightarrow True

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

1 - 2 sin (θ) = cos (θ)

plug in

θ = 2 π 1

1 - 2 sin (2 π 1) = cos (2 π 1)

Refine

1 = 1

\Rightarrow True

Check the solution

π + 2 πn :

False

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

1 - 2 sin (θ) = cos (θ)

plug in

θ = π + 2 π 1

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

Refine

1 = - 1

\Rightarrow False

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

Show solutions in decimal form

θ = π - 0.92729 \dots + 2 πn, θ = 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 1-2sin(θ)=cos(θ) ?
The general solution for 1-2sin(θ)=cos(θ) is θ=pi-0.92729…+2pin,θ=2pin