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

The general solution for 0=2sin^2(θ)+sqrt(3)sin(θ) is θ=2pin,θ=pi+2pin,θ=(4pi)/3+2pin,θ=(5pi)/3+2pin

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0=2sin^2(θ)+sqrt(3)sin(θ)

Solution

0 = 2 sin^{2} (θ) + 3 sin (θ)

Solution

θ = 2 πn, θ = π + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

0 = 2 sin^{2} (θ) + 3 sin (θ)

Switch sides

2 sin^{2} (θ) + 3 sin (θ) = 0

Solve by substitution

2 sin^{2} (θ) + 3 sin (θ) = 0

Let:

sin (θ) = u

2 u^{2} + 3 u = 0

2 u^{2} + 3 u = 0 : u = 0, u = - \frac{3}{2}

2 u^{2} + 3 u = 0

Solve with the quadratic formula

2 u^{2} + 3 u = 0

Quadratic Equation Formula:

For

a = 2, b = 3, c = 0

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

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

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

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

(3)^{2} = 3

(3)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Cancel the common factor:

2

= 1

= 3

4 \cdot 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 3 - 0

Subtract the numbers:

3 - 0 = 3

= 3

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

Separate the solutions

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

u = \frac{- 3 + 3}{2 \cdot 2} : 0

\frac{- 3 + 3}{2 \cdot 2}

Add similar elements:

- 3 + 3 = 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

u = \frac{- 3 - 3}{2 \cdot 2} : - \frac{3}{2}

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

Add similar elements:

- 3 - 3 = - 23

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 3}{4}

Apply the fraction rule:

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

= - \frac{2 3}{4}

Cancel the common factor:

2

= - \frac{3}{2}

The solutions to the quadratic equation are:

u = 0, u = - \frac{3}{2}

Substitute back

u = sin (θ)

sin (θ) = 0, sin (θ) = - \frac{3}{2}

sin (θ) = 0, sin (θ) = - \frac{3}{2}

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

sin (θ) = - \frac{3}{2} : θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

sin (θ) = - \frac{3}{2}

General solutions for

sin (θ) = - \frac{3}{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{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 0=2sin^2(θ)+sqrt(3)sin(θ) ?
The general solution for 0=2sin^2(θ)+sqrt(3)sin(θ) is θ=2pin,θ=pi+2pin,θ=(4pi)/3+2pin,θ=(5pi)/3+2pin