English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2sin(x)+2)/(cos(x))<= 0

Solution

\frac{2 sin ( x ) + 2}{cos ( x )} \leq 0

Solution

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

Interval Notation

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

Decimal Notation

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

Solution steps

\frac{2 sin ( x ) + 2}{cos ( x )} \leq 0

Periodicity of

\frac{2 sin ( x ) + 2}{cos ( x )} : 2 π

\frac{2 sin ( x ) + 2}{cos ( x )}

is composed of the following functions and periods:

sin (x)

with periodicity of

2 π

The compound periodicity is:

= 2 π

Find the zeroes and undifined points of

\frac{2 sin ( x ) + 2}{cos ( x )}

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

\frac{2 sin ( x ) + 2}{cos ( x )} = 0

\frac{2 sin ( x ) + 2}{cos ( x )} = 0, 0 \leq x < 2 π :

No Solution for

x \in R

\frac{2 sin ( x ) + 2}{cos ( x )} = 0, 0 \leq x < 2 π

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 sin (x) + 2 = 0

Move

2

to the right side

2 sin (x) + 2 = 0

Subtract

2

from both sides

2 sin (x) + 2 - 2 = 0 - 2

Simplify

2 sin (x) = - 2

2 sin (x) = - 2

Divide both sides by

2

2 sin (x) = - 2

Divide both sides by

2

\frac{2 sin ( x )}{2} = \frac{- 2}{2}

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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{3 π}{2}

Since the equation is undefined for:

\frac{3 π}{2}

No Solution for x \in R

Find the undefined points:

x = \frac{π}{2}, x = \frac{3 π}{2}

Find the zeros of the denominator

cos (x) = 0

General solutions for

cos (x) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

\frac{π}{2}, \frac{3 π}{2}

Identify the intervals

0 < x < \frac{π}{2}, \frac{π}{2} < x < \frac{3 π}{2}, \frac{3 π}{2} < x < 2 π

Summarize in a table:

2 sin (x) + 2 cos (x) \frac{2 s i n ( x ) + 2}{c o s ( x )} x = 0 + + + 0 < x < \frac{π}{2} + + + x = \frac{π}{2} + 0 Undefined \frac{π}{2} < x < \frac{3 π}{2} + - - x = \frac{3 π}{2} 00 Undefined \frac{3 π}{2} < x < 2 π + + + x = 2 π + + +

Identify the intervals that satisfy the required condition:

\leq 0

\frac{π}{2} < x < \frac{3 π}{2}

Apply the periodicity of

\frac{2 sin ( x ) + 2}{cos ( x )}

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

Popular Examples

sin(x)0<2pi

sin (x) 0 < 2 π

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

sin (x) - cos (x) - 1 \geq 0

sin(2x)>= 1/2

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

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

2 sin^{2} (x) - 1 \leq 0, 0 \leq x \leq 2 π

cos(2x)<1+sin(x)

cos (2 x) < 1 + sin (x)