What is the general solution for tan(x)(sin(x)-1)=0 ?

The general solution for tan(x)(sin(x)-1)=0 is x=pin

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

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

x = πn

Degrees

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

Solution steps

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

Solving each part separately

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

tan (x) = 0 : x = πn

tan (x) = 0

General solutions for

tan (x) = 0

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

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

sin (x) - 1 = 0

Move

1

to the right side

sin (x) - 1 = 0

Add

1

to both sides

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

Simplify

sin (x) = 1

sin (x) = 1

General solutions for

sin (x) = 1

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 = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = πn, x = \frac{π}{2} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn

x = πn

3 sin (x) + 1 = 0

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

tan (2 θ) = 2.93

cot (θ + 1) csc (θ - 1) = 0

sin (a) = 0.9455

What is the general solution for tan(x)(sin(x)-1)=0 ?
The general solution for tan(x)(sin(x)-1)=0 is x=pin