What is the general solution for sin(x)cot(x)-sin(x)=0 ?

The general solution for sin(x)cot(x)-sin(x)=0 is x= pi/4+pin

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sin(x)cot(x)-sin(x)=0

sin (x) cot (x) - sin (x) = 0

x = \frac{π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

sin (x) cot (x) - sin (x) = 0

Factor

sin (x) cot (x) - sin (x) : sin (x) (cot (x) - 1)

sin (x) cot (x) - sin (x)

Factor out common term

sin (x)

= sin (x) (cot (x) - 1)

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

Solving each part separately

sin (x) = 0 or cot (x) - 1 = 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

cot (x) - 1 = 0 : x = \frac{π}{4} + πn

cot (x) - 1 = 0

Move

1

to the right side

cot (x) - 1 = 0

Add

1

to both sides

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

Simplify

cot (x) = 1

cot (x) = 1

General solutions for

cot (x) = 1

cot (x)

periodicity table with

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + πn

Since the equation is undefined for:

2 πn, π + 2 πn

x = \frac{π}{4} + πn

tan (θ) = \frac{1}{7}

2 tan (x) + cos (x) = 0

cos (2 x) + sec (2 x) = 11

cos^{2} (t) = \frac{3}{4}

5 tan (x) - 4 = 0

What is the general solution for sin(x)cot(x)-sin(x)=0 ?
The general solution for sin(x)cot(x)-sin(x)=0 is x= pi/4+pin