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

The general solution for 5sin^2(θ)=2sin(θ) 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)

Solution

5 sin^{2} (θ) = 2 sin (θ)

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 = 156.42182 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sin^{2} (θ) = 2 sin (θ)

Solve by substitution

5 sin^{2} (θ) = 2 sin (θ)

Let:

sin (θ) = u

5 u^{2} = 2 u

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

5 u^{2} = 2 u

Move

2 u

to the left side

5 u^{2} = 2 u

Subtract

2 u

from both sides

5 u^{2} - 2 u = 2 u - 2 u

Simplify

5 u^{2} - 2 u = 0

5 u^{2} - 2 u = 0

Factor

5 u^{2} - 2 u : u (5 u - 2)

5 u^{2} - 2 u

Apply exponent rule:

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

u^{2} = uu

= 5 uu - 2 u

Factor out common term

u

= u (5 u - 2)

u (5 u - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or 5 u - 2 = 0

Solve

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

5 u - 2 = 0

Move

2

to the right side

5 u - 2 = 0

Add

2

to both sides

5 u - 2 + 2 = 0 + 2

Simplify

5 u = 2

5 u = 2

Divide both sides by

5

5 u = 2

Divide both sides by

5

\frac{5 u}{5} = \frac{2}{5}

Simplify

u = \frac{2}{5}

u = \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

Popular Examples

1+cos(x)=0

1 + cos (x) = 0

tan^2(θ)= 1/3

tan^{2} (θ) = \frac{1}{3}

sqrt(3)cot(x)-1=0

3 cot (x) - 1 = 0

cos(x)+1=-cos(x)

cos (x) + 1 = - cos (x)

sin(2θ)=-1

sin (2 θ) = - 1

Frequently Asked Questions (FAQ)

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