What is the general solution for 4sin^2(r)-sin(r)=0 ?

The general solution for 4sin^2(r)-sin(r)=0 is r=0.25268…+2pin,r=pi-0.25268…+2pin,r=2pin,r=pi+2pin

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4sin^2(r)-sin(r)=0

Solution

4 sin^{2} (r) - sin (r) = 0

Solution

r = 0.25268 \dots + 2 πn, r = π - 0.25268 \dots + 2 πn, r = 2 πn, r = π + 2 πn

Degrees

r = 14.47751 \dots^{\circ} + 36 0^{\circ} n, r = 165.52248 \dots^{\circ} + 36 0^{\circ} n, r = 0^{\circ} + 36 0^{\circ} n, r = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

4 sin^{2} (r) - sin (r) = 0

Solve by substitution

4 sin^{2} (r) - sin (r) = 0

Let:

sin (r) = u

4 u^{2} - u = 0

4 u^{2} - u = 0 : u = \frac{1}{4}, u = 0

4 u^{2} - u = 0

Solve with the quadratic formula

4 u^{2} - u = 0

Quadratic Equation Formula:

For

a = 4, b = - 1, c = 0

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

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

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

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

4 \cdot 4 \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 4}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 1}{2 \cdot 4}, u_{2} = \frac{- ( - 1 ) - 1}{2 \cdot 4}

u = \frac{- ( - 1 ) + 1}{2 \cdot 4} : \frac{1}{4}

\frac{- ( - 1 ) + 1}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{1 + 1}{2 \cdot 4}

Add the numbers:

1 + 1 = 2

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2}{8}

Cancel the common factor:

2

= \frac{1}{4}

u = \frac{- ( - 1 ) - 1}{2 \cdot 4} : 0

\frac{- ( - 1 ) - 1}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{1 - 1}{2 \cdot 4}

Subtract the numbers:

1 - 1 = 0

= \frac{0}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{0}{8}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = \frac{1}{4}, u = 0

Substitute back

u = sin (r)

sin (r) = \frac{1}{4}, sin (r) = 0

sin (r) = \frac{1}{4}, sin (r) = 0

sin (r) = \frac{1}{4} : r = arcsin (\frac{1}{4}) + 2 πn, r = π - arcsin (\frac{1}{4}) + 2 πn

sin (r) = \frac{1}{4}

Apply trig inverse properties

sin (r) = \frac{1}{4}

General solutions for

sin (r) = \frac{1}{4}

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

r = arcsin (\frac{1}{4}) + 2 πn, r = π - arcsin (\frac{1}{4}) + 2 πn

r = arcsin (\frac{1}{4}) + 2 πn, r = π - arcsin (\frac{1}{4}) + 2 πn

sin (r) = 0 : r = 2 πn, r = π + 2 πn

sin (r) = 0

General solutions for

sin (r) = 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}

r = 0 + 2 πn, r = π + 2 πn

r = 0 + 2 πn, r = π + 2 πn

Solve

r = 0 + 2 πn : r = 2 πn

r = 0 + 2 πn

0 + 2 πn = 2 πn

r = 2 πn

r = 2 πn, r = π + 2 πn

Combine all the solutions

r = arcsin (\frac{1}{4}) + 2 πn, r = π - arcsin (\frac{1}{4}) + 2 πn, r = 2 πn, r = π + 2 πn

Show solutions in decimal form

r = 0.25268 \dots + 2 πn, r = π - 0.25268 \dots + 2 πn, r = 2 πn, r = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(r)-sin(r)=0 ?
The general solution for 4sin^2(r)-sin(r)=0 is r=0.25268…+2pin,r=pi-0.25268…+2pin,r=2pin,r=pi+2pin