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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1-sin(θ)2=(1-sin(θ))(1-sin(θ))

Solution

1 - sin (θ) 2 = (1 - sin (θ)) (1 - sin (θ))

Solution

θ = 2 πn, θ = π + 2 πn

Degrees

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

Solution steps

1 - sin (θ) \cdot 2 = (1 - sin (θ)) (1 - sin (θ))

Solve by substitution

1 - sin (θ) \cdot 2 = (1 - sin (θ)) (1 - sin (θ))

Let:

sin (θ) = u

1 - u \cdot 2 = (1 - u) (1 - u)

1 - u \cdot 2 = (1 - u) (1 - u) : u = 0

1 - u \cdot 2 = (1 - u) (1 - u)

Expand

(1 - u) (1 - u) : 1 - 2 u + u^{2}

(1 - u) (1 - u)

Apply exponent rule:

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

(1 - u) (1 - u) = (1 - u)^{1 + 1}

= (1 - u)^{1 + 1}

Add the numbers:

1 + 1 = 2

= (1 - u)^{2}

Apply Perfect Square Formula:

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

a = 1, b = u

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

Simplify

1^{2} - 2 \cdot 1 \cdot u + u^{2} : 1 - 2 u + u^{2}

1^{2} - 2 \cdot 1 \cdot u + u^{2}

Apply rule

1^{a} = 1

1^{2} = 1

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

Multiply the numbers:

2 \cdot 1 = 2

= 1 - 2 u + u^{2}

= 1 - 2 u + u^{2}

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

Switch sides

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

- 2 u + u^{2} = - 2 u

- 2 u + u^{2} = - 2 u

Move

2 u

to the left side

- 2 u + u^{2} = - 2 u

Add

2 u

to both sides

- 2 u + u^{2} + 2 u = - 2 u + 2 u

Simplify

u^{2} = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

Substitute back

u = sin (θ)

sin (θ) = 0

sin (θ) = 0

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 = 2 πn, θ = π + 2 πn

Popular Examples

cos(x)= 22/24

cos (x) = \frac{22}{24}

sin(x)=(1.2)/(1.6)

sin (x) = \frac{1.2}{1.6}

2cos^2(θ)+3cos(θ)=-1

2 cos^{2} (θ) + 3 cos (θ) = - 1

sin(x)=(14.9)/(19.9725)

sin (x) = \frac{14.9}{19.9725}

sin(θ)= 1/2 ,0<= θ<= 4pi

sin (θ) = \frac{1}{2}, 0 \leq θ \leq 4 π

Frequently Asked Questions (FAQ)

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