What is the general solution for sin^2(θ)-3sin(θ)=0 ?

The general solution for sin^2(θ)-3sin(θ)=0 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 >

sin^2(θ)-3sin(θ)=0

Solution

sin^{2} (θ) - 3 sin (θ) = 0

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

sin^{2} (θ) - 3 sin (θ) = 0

Solve by substitution

sin^{2} (θ) - 3 sin (θ) = 0

Let:

sin (θ) = u

u^{2} - 3 u = 0

u^{2} - 3 u = 0 : u = 3, u = 0

u^{2} - 3 u = 0

Solve with the quadratic formula

u^{2} - 3 u = 0

Quadratic Equation Formula:

For

a = 1, b = - 3, c = 0

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot 0 = 3

(- 3)^{2} - 4 \cdot 1 \cdot 0

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 3^{2} - 0

3^{2} - 0 = 3^{2}

= 3^{2}

Apply radical rule:

assuming

a \geq 0

= 3

u_{1, 2} = \frac{- ( - 3 ) \pm 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 3}{2 \cdot 1}, u_{2} = \frac{- ( - 3 ) - 3}{2 \cdot 1}

u = \frac{- ( - 3 ) + 3}{2 \cdot 1} : 3

\frac{- ( - 3 ) + 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 + 3}{2 \cdot 1}

Add the numbers:

3 + 3 = 6

= \frac{6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

u = \frac{- ( - 3 ) - 3}{2 \cdot 1} : 0

\frac{- ( - 3 ) - 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 - 3}{2 \cdot 1}

Subtract the numbers:

3 - 3 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = 3, u = 0

Substitute back

u = sin (θ)

sin (θ) = 3, sin (θ) = 0

sin (θ) = 3, sin (θ) = 0

sin (θ) = 3 :

No Solution

sin (θ) = 3

- 1 \leq sin (x) \leq 1

No Solution

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^2(x)= 1/5

cos^{2} (x) = \frac{1}{5}

-3cos(2θ)+5=7sin(θ)+5

- 3 cos (2 θ) + 5 = 7 sin (θ) + 5

3+3cos(x)=2sin^2(x)

3 + 3 cos (x) = 2 sin^{2} (x)

cos(θ)=-sin(θ)

cos (θ) = - sin (θ)

sqrt(2)csc(x)+2=4

2 csc (x) + 2 = 4

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(θ)-3sin(θ)=0 ?
The general solution for sin^2(θ)-3sin(θ)=0 is θ=2pin,θ=pi+2pin