What is the general solution for 5sin^2(θ)+2sin(θ)=0 ?

The general solution for 5sin^2(θ)+2sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=-0.41151…+2pin,θ=pi+0.41151…+2pin

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Popular >

5sin^2(θ)+2sin(θ)=0

Solution

5 sin^{2} (θ) + 2 sin (θ) = 0

Solution

θ = 2 πn, θ = π + 2 πn, θ = - 0.41151 \dots + 2 πn, θ = π + 0.41151 \dots + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = - 23.57817 \dots^{\circ} + 36 0^{\circ} n, θ = 203.57817 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sin^{2} (θ) + 2 sin (θ) = 0

Solve by substitution

5 sin^{2} (θ) + 2 sin (θ) = 0

Let:

sin (θ) = u

5 u^{2} + 2 u = 0

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

5 u^{2} + 2 u = 0

Solve with the quadratic formula

5 u^{2} + 2 u = 0

Quadratic Equation Formula:

For

a = 5, b = 2, c = 0

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

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

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

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

Apply rule

0 \cdot a = 0

= 2^{2} - 0

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

= 2^{2}

Apply radical rule:

n a^{n} = a,

assuming

a \geq 0

= 2

u_{1, 2} = \frac{- 2 \pm 2}{2 \cdot 5}

Separate the solutions

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

u = \frac{- 2 + 2}{2 \cdot 5} : 0

\frac{- 2 + 2}{2 \cdot 5}

Add/Subtract the numbers:

- 2 + 2 = 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

u = \frac{- 2 - 2}{2 \cdot 5} : - \frac{2}{5}

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

Subtract the numbers:

- 2 - 2 = - 4

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 4}{10}

Apply the fraction rule:

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

= - \frac{4}{10}

Cancel the common factor:

2

= - \frac{2}{5}

The solutions to the quadratic equation are:

u = 0, u = - \frac{2}{5}

Substitute back

u = sin (θ)

sin (θ) = 0, sin (θ) = - \frac{2}{5}

sin (θ) = 0, sin (θ) = - \frac{2}{5}

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

sin (θ) = - \frac{2}{5}

Apply trig inverse properties

sin (θ) = - \frac{2}{5}

General solutions for

sin (θ) = - \frac{2}{5}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = - 0.41151 \dots + 2 πn, θ = π + 0.41151 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 5sin^2(θ)+2sin(θ)=0 ?
The general solution for 5sin^2(θ)+2sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=-0.41151…+2pin,θ=pi+0.41151…+2pin