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

The general solution for (tan(x)+1)(sin(x)-1)=0 is x=(3pi)/4+pin

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

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

x = \frac{3 π}{4} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Solving each part separately

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

tan (x) + 1 = 0 : x = \frac{3 π}{4} + πn

tan (x) + 1 = 0

Move

1

to the right side

tan (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (x) = - 1

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 = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + 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 = \frac{3 π}{4} + πn, x = \frac{π}{2} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn

x = \frac{3 π}{4} + πn

5 cos (a) + 8 = 3 cos (a) + 6

cos (x) = \frac{- 3}{4}

3 cot^{2} (x) - 9 = 0

\frac{2}{5} = cos (θ)

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

What is the general solution for (tan(x)+1)(sin(x)-1)=0 ?
The general solution for (tan(x)+1)(sin(x)-1)=0 is x=(3pi)/4+pin