What is the general solution for sin^3(x)cos(x)-sin^2(x)=0 ?

The general solution for sin^3(x)cos(x)-sin^2(x)=0 is x=2pin,x=pi+2pin

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Popular >

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

Solution

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

Solution

x = 2 πn, x = π + 2 πn

Degrees

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

Solution steps

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

Factor

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

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

Apply exponent rule:

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

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

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

Factor out common term

sin^{2} (x)

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

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

Solving each part separately

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

sin^{2} (x) = 0 : x = 2 πn, x = π + 2 πn

sin^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

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

sin (x) cos (x) - 1 = 0 :

No Solution

sin (x) cos (x) - 1 = 0

Rewrite using trig identities

sin (x) cos (x) - 1

Use the Double Angle identity:

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

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= - 1 + \frac{sin ( 2 x )}{2}

- 1 + \frac{sin ( 2 x )}{2} = 0

Move

1

to the right side

- 1 + \frac{sin ( 2 x )}{2} = 0

Add

1

to both sides

- 1 + \frac{sin ( 2 x )}{2} + 1 = 0 + 1

Simplify

\frac{sin ( 2 x )}{2} = 1

\frac{sin ( 2 x )}{2} = 1

Multiply both sides by

2

\frac{sin ( 2 x )}{2} = 1

Multiply both sides by

2

\frac{2 sin ( 2 x )}{2} = 1 \cdot 2

Simplify

sin (2 x) = 2

sin (2 x) = 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

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sin^{2} (x) + sin^{4} (x) = 0

cos^2(45-a)-sin^2(45-a)=sin^2(a)

cos^{2} (4 5^{\circ} - a) - sin^{2} (4 5^{\circ} - a) = sin^{2} (a)

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

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

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

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

2cos(x)=-3

2 cos (x) = - 3

Frequently Asked Questions (FAQ)

What is the general solution for sin^3(x)cos(x)-sin^2(x)=0 ?
The general solution for sin^3(x)cos(x)-sin^2(x)=0 is x=2pin,x=pi+2pin