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2sin(x)-1<0,-2pi<= x<= 0

Solution

2 sin (x) - 1 < 0, - 2 π \leq x \leq 0

- 2 π \leq x < \frac{π}{6} - 2 π or - \frac{7 π}{6} < x \leq 0

Interval Notation

[- 2 π, \frac{π}{6} - 2 π) \cup (- \frac{7 π}{6}, 0]

Decimal

- 6.28318 \dots \leq x < - 5.75958 \dots or - 3.66519 \dots < x \leq 0

Solution steps

2 sin (x) - 1 < 0, - 2 π \leq x \leq 0

Move

1

to the right side

2 sin (x) - 1 < 0

Add

1

to both sides

2 sin (x) - 1 + 1 < 0 + 1

Simplify

2 sin (x) < 1

2 sin (x) < 1

Divide both sides by

2

2 sin (x) < 1

Divide both sides by

2

\frac{2 sin ( x )}{2} < \frac{1}{2}

Simplify

sin (x) < \frac{1}{2}

sin (x) < \frac{1}{2}

For

sin (x) < a

, if

- 1 < a \leq 1

then

- π - arcsin (a) + 2 πn < x < arcsin (a) + 2 πn

- π - arcsin (\frac{1}{2}) + 2 πn < x < arcsin (\frac{1}{2}) + 2 πn

Simplify

- π - arcsin (\frac{1}{2}) : - \frac{7 π}{6}

- π - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

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 - \frac{π}{6}

Simplify

- π - \frac{π}{6}

Convert element to fraction:

π = \frac{π 6}{6}

= - \frac{π 6}{6} - \frac{π}{6}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- π 6 - π}{6}

Add similar elements:

- 6 π - π = - 7 π

= \frac{- 7 π}{6}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{7 π}{6}

= - \frac{7 π}{6}

Simplify

arcsin (\frac{1}{2}) : \frac{π}{6}

arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

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}

= \frac{π}{6}

- \frac{7 π}{6} + 2 πn < x < \frac{π}{6} + 2 πn

Combine the intervals

- \frac{7 π}{6} + 2 πn < x < \frac{π}{6} + 2 πn and - 2 π \leq x \leq 0

- 2 π \leq x < \frac{π}{6} - 2 π or - \frac{7 π}{6} < x \leq 0

\frac{cos ( x )}{1 + cos ( 2 x )} < 0

\frac{tan ( x ) - tan ^{2} ( x )}{2 sin ( x ) - 1} < 0

sin (x) - cos (x) > 1

sin (y) > 0

2 sin (x) + 1 \geq 0