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solvefor x,sin(((x))/(cos(y)))>0

Solution

solve for

x, sin (\frac{( x )}{cos ( y )}) > 0

2 πn cos (y) < x < π cos (y) + 2 πn cos (y)

Solution steps

sin (\frac{x}{cos ( y )}) > 0

For

sin (x) > a

, if

- 1 \leq a < 1

then

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

arcsin (0) + 2 πn < \frac{x}{cos ( y )} < π - arcsin (0) + 2 πn

a < u < b

then

a < u and u < b

arcsin (0) + 2 πn < \frac{x}{cos ( y )} and \frac{x}{cos ( y )} < π - arcsin (0) + 2 πn

arcsin (0) + 2 πn < \frac{x}{cos ( y )} : x > 2 πn cos (y)

arcsin (0) + 2 πn < \frac{x}{cos ( y )}

Switch sides

\frac{x}{cos ( y )} > arcsin (0) + 2 πn

Simplify

arcsin (0) + 2 πn : 2 πn

arcsin (0) + 2 πn

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 + 2 πn

0 + 2 πn = 2 πn

= 2 πn

\frac{x}{cos ( y )} > 2 πn

Multiply both sides by

cos (y)

\frac{x}{cos ( y )} > 2 πn

Multiply both sides by

cos (y)

\frac{x cos ( y )}{cos ( y )} > 2 πn cos (y)

Simplify

x > 2 πn cos (y)

x > 2 πn cos (y)

\frac{x}{cos ( y )} < π - arcsin (0) + 2 πn : x < π cos (y) + 2 πn cos (y)

\frac{x}{cos ( y )} < π - arcsin (0) + 2 πn

Simplify

π - arcsin (0) + 2 πn : π + 2 πn

π - arcsin (0) + 2 πn

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 + 2 πn

π - 0 + 2 πn = π + 2 πn

= π + 2 πn

\frac{x}{cos ( y )} < π + 2 πn

Multiply both sides by

cos (y)

\frac{x}{cos ( y )} < π + 2 πn

Multiply both sides by

cos (y)

\frac{x cos ( y )}{cos ( y )} < π cos (y) + 2 πn cos (y)

Simplify

x < π cos (y) + 2 πn cos (y)

x < π cos (y) + 2 πn cos (y)

Combine the intervals

x > 2 πn cos (y) and x < π cos (y) + 2 πn cos (y)

Merge Overlapping Intervals

2 πn cos (y) < x < π cos (y) + 2 πn cos (y)

cos (\frac{x}{2}) \geq - \frac{1}{2}

- 5 (sin (x) + cos (x)) > 0

- sin (x) \leq 1

13 > 2.5 cos ((\frac{2 π}{365}) (x + 9)) + 12

\frac{9 cos ( t ) - \frac{1}{2}}{4} \geq 4