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2sin(x)-sec(x)>0

Solution

2 sin (x) - sec (x) > 0

Solution

\frac{π}{2} + 2 πn < x < \frac{5 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn < x < \frac{3 π}{2} + 2 πn

Interval Notation

(\frac{π}{2} + 2 πn, \frac{5 π}{4} + 2 πn) \cup (\frac{5 π}{4} + 2 πn, \frac{3 π}{2} + 2 πn)

Decimal Notation

1.57079 \dots + 2 πn < x < 3.92699 \dots + 2 πn or 3.92699 \dots + 2 πn < x < 4.71238 \dots + 2 πn

Solution steps

2 sin (x) - sec (x) > 0

Periodicity of

2 sin (x) - sec (x) : 2 π

The compound periodicity of the sum of periodic functions is the least common multiplier of the periods

2 sin (x), sec (x)

Periodicity of

2 sin (x) : 2 π

Periodicity of

a \cdot sin (b x + c) + d = \frac{periodicity of sin ( x )}{∣ b ∣}

Periodicity of

sin (x)

2 π

= \frac{2 π}{∣ 1 ∣}

Simplify

= 2 π

Periodicity of

sec (x) : 2 π

Periodicity of

sec (x)

2 π

= 2 π

Combine periods:

2 π, 2 π

= 2 π

Express with sin, cos

2 sin (x) - sec (x) > 0

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Find the zeroes and undifined points of

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

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

\frac{2 sin ( x ) cos ( x ) - 1}{cos ( x )} = 0, 0 \leq x < 2 π : x = \frac{π}{4}, x = \frac{5 π}{4}

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

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

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

Rewrite using trig identities

2 sin (x) cos (x) - 1

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= - 1 + sin (2 x)

- 1 + sin (2 x) = 0

Move

1

to the right side

- 1 + sin (2 x) = 0

Add

1

to both sides

- 1 + sin (2 x) + 1 = 0 + 1

Simplify

sin (2 x) = 1

sin (2 x) = 1

General solutions for

sin (2 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}

2 x = \frac{π}{2} + 2 πn

2 x = \frac{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

x = \frac{π}{4} + πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{4}, x = \frac{5 π}{4}

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}{4}, \frac{π}{2}, \frac{5 π}{4}, \frac{3 π}{2}

Identify the intervals

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

Summarize in a table:

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

> 0

\frac{π}{2} < x < \frac{5 π}{4} or \frac{5 π}{4} < x < \frac{3 π}{2}

Apply the periodicity of

2 sin (x) - sec (x)

\frac{π}{2} + 2 πn < x < \frac{5 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn < x < \frac{3 π}{2} + 2 πn

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