What is the general solution for sin(x)cos^2(x)=sin(x),0<= x<= 2pi ?

Q: What is the general solution for sin(x)cos^2(x)=sin(x),0<= x<= 2pi ?

The general solution for sin(x)cos^2(x)=sin(x),0<= x<= 2pi is x=0,x=pi,x=2pi

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sin(x)cos^2(x)=sin(x),0<= x<= 2pi

Solution

sin (x) cos^{2} (x) = sin (x), 0 \leq x \leq 2 π

Solution

x = 0, x = π, x = 2 π

Degrees

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

Solution steps

sin (x) cos^{2} (x) = sin (x), 0 \leq x \leq 2 π

Subtract

sin (x)

from both sides

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

Factor

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

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

Factor out common term

sin (x)

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

Factor

cos^{2} (x) - 1 : (cos (x) + 1) (cos (x) - 1)

cos^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

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

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

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

Solving each part separately

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

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

sin (x) = 0, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

cos (x) + 1 = 0, 0 \leq x \leq 2 π : x = π

cos (x) + 1 = 0, 0 \leq x \leq 2 π

Move

1

to the right side

cos (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

cos (x) = - 1

cos (x) = - 1

General solutions for

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = π

cos (x) - 1 = 0, 0 \leq x \leq 2 π : x = 0, x = 2 π

cos (x) - 1 = 0, 0 \leq x \leq 2 π

Move

1

to the right side

cos (x) - 1 = 0

Add

1

to both sides

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

Simplify

cos (x) = 1

cos (x) = 1

General solutions for

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 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

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = 2 π

Combine all the solutions

x = 0, x = π, x = 2 π

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x)cos^2(x)=sin(x),0<= x<= 2pi ?
The general solution for sin(x)cos^2(x)=sin(x),0<= x<= 2pi is x=0,x=pi,x=2pi