What is the general solution for -sin(θ)+2sin^2(θ)=0 ?

The general solution for -sin(θ)+2sin^2(θ)=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin,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 >

-sin(θ)+2sin^2(θ)=0

Solution

- sin (θ) + 2 sin^{2} (θ) = 0

Solution

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = 2 πn, θ = π + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

- sin (θ) + 2 sin^{2} (θ) = 0

Solve by substitution

- sin (θ) + 2 sin^{2} (θ) = 0

Let:

sin (θ) = u

- u + 2 u^{2} = 0

- u + 2 u^{2} = 0 : u = \frac{1}{2}, u = 0

- u + 2 u^{2} = 0

Write in the standard form

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

2 u^{2} - u = 0

Solve with the quadratic formula

2 u^{2} - u = 0

Quadratic Equation Formula:

For

a = 2, b = - 1, c = 0

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

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

(- 1)^{2} - 4 \cdot 2 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 2 \cdot 0

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

\frac{- ( - 1 ) + 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 1}{2 \cdot 2}

Add the numbers:

1 + 1 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- ( - 1 ) - 1}{2 \cdot 2} : 0

\frac{- ( - 1 ) - 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 - 1}{2 \cdot 2}

Subtract the numbers:

1 - 1 = 0

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = 0

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{2}, sin (θ) = 0

sin (θ) = \frac{1}{2}, sin (θ) = 0

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

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 = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 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 = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = 2 πn, θ = π + 2 πn

Popular Examples

-sec(x)+sqrt(2)sec(x)cos(x)=0

- sec (x) + 2 sec (x) cos (x) = 0

2tan(x)-3sin(x)=0

2 tan (x) - 3 sin (x) = 0

sin(x)+5=4

sin (x) + 5 = 4

-9cos(x)=0

- 9 cos (x) = 0

-3cos(x)=0

- 3 cos (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for -sin(θ)+2sin^2(θ)=0 ?
The general solution for -sin(θ)+2sin^2(θ)=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=2pin,θ=pi+2pin