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Popular Trigonometry >

(2sin(x)-1)/(3cos(x))<= 0

Solution

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

Solution

2 πn \leq x \leq \frac{π}{6} + 2 πn or \frac{π}{2} + 2 πn < x \leq \frac{5 π}{6} + 2 πn or \frac{3 π}{2} + 2 πn < x \leq 2 π + 2 πn

Interval Notation

[2 πn, \frac{π}{6} + 2 πn] \cup (\frac{π}{2} + 2 πn, \frac{5 π}{6} + 2 πn] \cup (\frac{3 π}{2} + 2 πn, 2 π + 2 πn]

Decimal

2 πn \leq x \leq 0.52359 \dots + 2 πn or 1.57079 \dots + 2 πn < x \leq 2.61799 \dots + 2 πn or 4.71238 \dots + 2 πn < x \leq 6.28318 \dots + 2 πn

Solution steps

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

Periodicity of

\frac{2 sin ( x ) - 1}{3 cos ( x )} : 2 π

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

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

\frac{2 sin ( x ) - 1}{3 cos ( x )} = 0, 0 \leq x < 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

\frac{2 sin ( x ) - 1}{3 cos ( x )} = 0, 0 \leq x < 2 π

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

2 sin (x) - 1 = 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}

General solutions for

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

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{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}

Find the undefined points:

x = \frac{π}{2}, x = \frac{3 π}{2}

Find the zeros of the denominator

3 cos (x) = 0

Divide both sides by

3

3 cos (x) = 0

Divide both sides by

3

\frac{3 cos ( x )}{3} = \frac{0}{3}

Simplify

cos (x) = 0

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}{6}, \frac{π}{2}, \frac{5 π}{6}, \frac{3 π}{2}

Identify the intervals

0 < x < \frac{π}{6}, \frac{π}{6} < x < \frac{π}{2}, \frac{π}{2} < x < \frac{5 π}{6}, \frac{5 π}{6} < x < \frac{3 π}{2}, \frac{3 π}{2} < x < 2 π

Summarize in a table:

2 sin (x) - 1 cos (x) \frac{2 s i n ( x ) - 1}{3 c o s ( x )} x = 0 - + - 0 < x < \frac{π}{6} - + - x = \frac{π}{6} 0 + 0 \frac{π}{6} < x < \frac{π}{2} + + + x = \frac{π}{2} + 0 Undefined \frac{π}{2} < x < \frac{5 π}{6} + - - x = \frac{5 π}{6} 0 - 0 \frac{5 π}{6} < x < \frac{3 π}{2} - - + x = \frac{3 π}{2} - 0 Undefined \frac{3 π}{2} < x < 2 π - + - x = 2 π - + -

Identify the intervals that satisfy the required condition:

\leq 0

x = 0 or 0 < x < \frac{π}{6} or x = \frac{π}{6} or \frac{π}{2} < x < \frac{5 π}{6} or x = \frac{5 π}{6} or \frac{3 π}{2} < x < 2 π or x = 2 π

Merge Overlapping Intervals

0 \leq x \leq \frac{π}{6} or \frac{π}{2} < x \leq \frac{5 π}{6} or \frac{3 π}{2} < x < 2 π or x = 2 π