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

The general solution for 5sin^2(x)=sin(x) is x=0.20135…+2pin,x=pi-0.20135…+2pin,x=2pin,x=pi+2pin

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Popular >

5sin^2(x)=sin(x)

Solution

5 sin^{2} (x) = sin (x)

Solution

x = 0.20135 \dots + 2 πn, x = π - 0.20135 \dots + 2 πn, x = 2 πn, x = π + 2 πn

Degrees

x = 11.53695 \dots^{\circ} + 36 0^{\circ} n, x = 168.46304 \dots^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

5 sin^{2} (x) = sin (x)

Solve by substitution

5 sin^{2} (x) = sin (x)

Let:

sin (x) = u

5 u^{2} = u

5 u^{2} = u : u = \frac{1}{5}, u = 0

5 u^{2} = u

Move

u

to the left side

5 u^{2} = u

Subtract

u

from both sides

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

Simplify

5 u^{2} - u = 0

5 u^{2} - u = 0

Solve with the quadratic formula

5 u^{2} - u = 0

Quadratic Equation Formula:

For

a = 5, b = - 1, c = 0

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

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

(- 1)^{2} - 4 \cdot 5 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 5 \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 5 \cdot 0 = 0

4 \cdot 5 \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 5}

Separate the solutions

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

u = \frac{- ( - 1 ) + 1}{2 \cdot 5} : \frac{1}{5}

\frac{- ( - 1 ) + 1}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{1 + 1}{2 \cdot 5}

Add the numbers:

1 + 1 = 2

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{2}{10}

Cancel the common factor:

2

= \frac{1}{5}

u = \frac{- ( - 1 ) - 1}{2 \cdot 5} : 0

\frac{- ( - 1 ) - 1}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{1 - 1}{2 \cdot 5}

Subtract the numbers:

1 - 1 = 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

The solutions to the quadratic equation are:

u = \frac{1}{5}, u = 0

Substitute back

u = sin (x)

sin (x) = \frac{1}{5}, sin (x) = 0

sin (x) = \frac{1}{5}, sin (x) = 0

sin (x) = \frac{1}{5} : x = arcsin (\frac{1}{5}) + 2 πn, x = π - arcsin (\frac{1}{5}) + 2 πn

sin (x) = \frac{1}{5}

Apply trig inverse properties

sin (x) = \frac{1}{5}

General solutions for

sin (x) = \frac{1}{5}

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

x = arcsin (\frac{1}{5}) + 2 πn, x = π - arcsin (\frac{1}{5}) + 2 πn

x = arcsin (\frac{1}{5}) + 2 πn, x = π - arcsin (\frac{1}{5}) + 2 πn

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = arcsin (\frac{1}{5}) + 2 πn, x = π - arcsin (\frac{1}{5}) + 2 πn, x = 2 πn, x = π + 2 πn

Show solutions in decimal form

x = 0.20135 \dots + 2 πn, x = π - 0.20135 \dots + 2 πn, x = 2 πn, x = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5sin^2(x)=sin(x) ?
The general solution for 5sin^2(x)=sin(x) is x=0.20135…+2pin,x=pi-0.20135…+2pin,x=2pin,x=pi+2pin