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Popular >

1-sin(x)>0

Solution

1 - sin (x) > 0

- \frac{3 π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

Interval Notation

(- \frac{3 π}{2} + 2 πn, \frac{π}{2} + 2 πn)

Decimal Notation

- 4.71238 \dots + 2 πn < x < 1.57079 \dots + 2 πn

Solution steps

1 - sin (x) > 0

Move

1

to the right side

1 - sin (x) > 0

Subtract

1

from both sides

1 - sin (x) - 1 > 0 - 1

Simplify

- sin (x) > - 1

- sin (x) > - 1

Multiply both sides by

- 1

- sin (x) > - 1

Multiply both sides by -1 (reverse the inequality)

(- sin (x)) (- 1) < (- 1) (- 1)

Simplify

sin (x) < 1

sin (x) < 1

For

sin (x) < a

, if

- 1 < a \leq 1

then

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

- π - arcsin (1) + 2 πn < x < arcsin (1) + 2 πn

Simplify

- π - arcsin (1) : - \frac{3 π}{2}

- π - arcsin (1)

Use the following trivial identity:

arcsin (1) = \frac{π}{2}

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}{2}

Simplify

- π - \frac{π}{2}

Convert element to fraction:

π = \frac{π 2}{2}

= - \frac{π 2}{2} - \frac{π}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 2 - π}{2}

Add similar elements:

- 2 π - π = - 3 π

= \frac{- 3 π}{2}

Apply the fraction rule:

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

= - \frac{3 π}{2}

= - \frac{3 π}{2}

Simplify

arcsin (1) : \frac{π}{2}

arcsin (1)

Use the following trivial identity:

arcsin (1) = \frac{π}{2}

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}{2}

- \frac{3 π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

2 + 3 sin (x) - cos (2 x) < 0

2 cos^{2} (X) \leq 1 - 2 cos (2 X)

tan (θ) - \frac{4}{5} cos (θ) > 0 csc (θ)

cot (x) > - \frac{1}{3}

(cos (x) - 1.5708) (cos (x) + 1.5708) \leq 0