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

The general solution for sin(x)-1=0,sin(x)-1=0 is No Solution for x\in\mathbb{R}

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

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

No Solution for x \in R

Solution steps

sin (x) - 1 = 0, 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

Solutions for the range

sin (x) - 1 = 0

No Solution for x \in R

23 cos (2 x) = - 3

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

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

e^{c o s (x)} sin^{2} (x) - e^{c o s (x)} cos (x) = 0

2 cot (x) + tan (x) = - 2

What is the general solution for sin(x)-1=0,sin(x)-1=0 ?
The general solution for sin(x)-1=0,sin(x)-1=0 is No Solution for x\in\mathbb{R}