sin(x)>= 0,(1-2sin(x))/(1+sin(x))>0

Solution

sin (x) \geq 0, \frac{1 - 2 sin ( x )}{1 + sin ( x )} > 0

2 πn \leq x \leq π + 2 πn

Interval Notation

[2 πn, π + 2 πn]

Decimal

2 πn \leq x \leq 3.14159 \dots + 2 πn

Solution steps

sin (x) \geq 0

For

sin (x) \geq a

, if

- 1 < a < 1

then

arcsin (a) + 2 πn \leq x \leq π - arcsin (a) + 2 πn

arcsin (0) + 2 πn \leq x \leq π - arcsin (0) + 2 πn

Simplify

arcsin (0) : 0

arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= 0

Simplify

π - arcsin (0) : π

π - arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= π - 0

π - 0 = π

= π

0 + 2 πn \leq x \leq π + 2 πn

Simplify

2 πn \leq x \leq π + 2 πn