What is the general solution for sin(x)-sin(2x)=sin^3(x) ?

The general solution for sin(x)-sin(2x)=sin^3(x) is x=2pin,x=pi+2pin,x= pi/2+2pin,x=(3pi)/2+2pin

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Popular >

sin(x)-sin(2x)=sin^3(x)

Solution

sin (x) - sin (2 x) = sin^{3} (x)

Solution

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (x) - sin (2 x) = sin^{3} (x)

Subtract

sin^{3} (x)

from both sides

sin (x) - sin (2 x) - sin^{3} (x) = 0

Rewrite using trig identities

- sin (2 x) + sin (x) - sin^{3} (x)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= - 2 sin (x) cos (x) + sin (x) - sin^{3} (x)

sin (x) - sin^{3} (x) - 2 cos (x) sin (x) = 0

Factor

sin (x) - sin^{3} (x) - 2 cos (x) sin (x) : sin (x) (1 - sin^{2} (x) - 2 cos (x))

sin (x) - sin^{3} (x) - 2 cos (x) sin (x)

Apply exponent rule:

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

sin^{3} (x) = sin (x) sin^{2} (x)

= sin (x) - sin (x) sin^{2} (x) - 2 sin (x) cos (x)

Factor out common term

sin (x)

= sin (x) (1 - sin^{2} (x) - 2 cos (x))

sin (x) (1 - sin^{2} (x) - 2 cos (x)) = 0

Solving each part separately

sin (x) = 0 or 1 - sin^{2} (x) - 2 cos (x) = 0

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

1 - sin^{2} (x) - 2 cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

1 - sin^{2} (x) - 2 cos (x) = 0

Rewrite using trig identities

1 - sin^{2} (x) - 2 cos (x)

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - sin^{2} (x) = cos^{2} (x)

= - 2 cos (x) + cos^{2} (x)

cos^{2} (x) - 2 cos (x) = 0

Solve by substitution

cos^{2} (x) - 2 cos (x) = 0

Let:

cos (x) = u

u^{2} - 2 u = 0

u^{2} - 2 u = 0 : u = 2, u = 0

u^{2} - 2 u = 0

Solve with the quadratic formula

u^{2} - 2 u = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = 0

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

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

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

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

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

Apply rule

0 \cdot a = 0

= 2^{2} - 0

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

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

2 + 2 = 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

2 - 2 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = 2, u = 0

Substitute back

u = cos (x)

cos (x) = 2, cos (x) = 0

cos (x) = 2, cos (x) = 0

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Popular Examples

-sin^2(x)=cos(2x)

- sin^{2} (x) = cos (2 x)

sin(x)= 4/5 , pi/2 <= 0<pi,sin(2x)

sin (x) = \frac{4}{5}, \frac{π}{2} \leq 0 < π, sin (2 x)

2cos^2(θ)-3cos(θ)+1=0,(3pi)/2 <θ<2pi

2 cos^{2} (θ) - 3 cos (θ) + 1 = 0, \frac{3 π}{2} < θ < 2 π

-2cos^2(x)+cos(x)=0

- 2 cos^{2} (x) + cos (x) = 0

3cos(x)-2sin(2x)=0

3 cos (x) - 2 sin (2 x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for sin(x)-sin(2x)=sin^3(x) ?
The general solution for sin(x)-sin(2x)=sin^3(x) is x=2pin,x=pi+2pin,x= pi/2+2pin,x=(3pi)/2+2pin