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Popular Trigonometry >

(1+2sin(x))/((cos(2x)))>0

Solution

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

Solution

2 πn \leq x < \frac{π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < \frac{7 π}{6} + 2 πn or \frac{5 π}{4} + 2 πn < x < \frac{7 π}{4} + 2 πn or \frac{11 π}{6} + 2 πn < x \leq 2 π + 2 πn

Interval Notation

[2 πn, \frac{π}{4} + 2 πn) \cup (\frac{3 π}{4} + 2 πn, \frac{7 π}{6} + 2 πn) \cup (\frac{5 π}{4} + 2 πn, \frac{7 π}{4} + 2 πn) \cup (\frac{11 π}{6} + 2 πn, 2 π + 2 πn]

Decimal

2 πn \leq x < 0.78539 \dots + 2 πn or 2.35619 \dots + 2 πn < x < 3.66519 \dots + 2 πn or 3.92699 \dots + 2 πn < x < 5.49778 \dots + 2 πn or 5.75958 \dots + 2 πn < x \leq 6.28318 \dots + 2 πn

Solution steps

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

Use the following identity:

cos (2 x) = cos^{2} (x) - sin^{2} (x)

\frac{1 + 2 sin ( x )}{cos ^{2} ( x ) - sin ^{2} ( x )} > 0

Periodicity of

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

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

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

\frac{1 + 2 sin ( x )}{cos ^{2} ( x ) - sin ^{2} ( x )} = 0, 0 \leq x < 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

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

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

1 + 2 sin (x) = 0

Move

1

to the right side

1 + 2 sin (x) = 0

Subtract

1

from both sides

1 + 2 sin (x) - 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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

Find the undefined points:

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

Find the zeros of the denominator

cos^{2} (x) - sin^{2} (x) = 0

Rewrite using trig identities

cos^{2} (x) - sin^{2} (x)

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

= cos (2 x)

cos (2 x) = 0

General solutions for

cos (2 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}

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

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

Solve

2 x = \frac{π}{2} + 2 πn : x = \frac{π}{4} + πn

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Solve

2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{4} + πn

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

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

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

\frac{π}{4}, \frac{3 π}{4}, \frac{7 π}{6}, \frac{5 π}{4}, \frac{7 π}{4}, \frac{11 π}{6}

Identify the intervals

0 < x < \frac{π}{4}, \frac{π}{4} < x < \frac{3 π}{4}, \frac{3 π}{4} < x < \frac{7 π}{6}, \frac{7 π}{6} < x < \frac{5 π}{4}, \frac{5 π}{4} < x < \frac{7 π}{4}, \frac{7 π}{4} < x < \frac{11 π}{6}, \frac{11 π}{6} < x < 2 π

Summarize in a table:

1 + 2 sin (x) cos^{2} (x) - sin^{2} (x) \frac{1 + 2 s i n ( x )}{c o s ^{2} ( x ) - s i n ^{2} ( x )} x = 0 + + + 0 < x < \frac{π}{4} + + + x = \frac{π}{4} + 0 Undefined \frac{π}{4} < x < \frac{3 π}{4} + - - x = \frac{3 π}{4} + 0 Undefined \frac{3 π}{4} < x < \frac{7 π}{6} + + + x = \frac{7 π}{6} 0 + 0 \frac{7 π}{6} < x < \frac{5 π}{4} - + - x = \frac{5 π}{4} - 0 Undefined \frac{5 π}{4} < x < \frac{7 π}{4} - - + x = \frac{7 π}{4} - 0 Undefined \frac{7 π}{4} < x < \frac{11 π}{6} - + - x = \frac{11 π}{6} 0 + 0 \frac{11 π}{6} < x < 2 π + + + x = 2 π + + +

Identify the intervals that satisfy the required condition:

> 0

x = 0 or 0 < x < \frac{π}{4} or \frac{3 π}{4} < x < \frac{7 π}{6} or \frac{5 π}{4} < x < \frac{7 π}{4} or \frac{11 π}{6} < x < 2 π or x = 2 π

Merge Overlapping Intervals

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x < \frac{7 π}{6} or \frac{5 π}{4} < x < \frac{7 π}{4} or \frac{11 π}{6} < x < 2 π or x = 2 π