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Popular >

tan(x)*sin(x)> 1/(2cos(x))

Solution

tan (x) \cdot sin (x) > \frac{1}{2 cos ( x )}

Solution

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

Interval Notation

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

Decimal Notation

0.78539 \dots + 2 πn < x < 1.57079 \dots + 2 πn or 2.35619 \dots + 2 πn < x < 3.92699 \dots + 2 πn or 4.71238 \dots + 2 πn < x < 5.49778 \dots + 2 πn

Solution steps

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

Move

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

to the left side

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

Subtract

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

from both sides

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

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

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

Periodicity of

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

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

tan (x) sin (x), \frac{1}{2 cos ( x )}

Periodicity of

tan (x) sin (x) : 2 π

tan (x) sin (x)

is composed of the following functions and periods:

tan (x)

with periodicity of

π

The compound periodicity is:

2 π

Periodicity of

\frac{1}{2 cos ( x )} : 2 π

Periodicity of

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

Periodicity of

cos (x)

2 π

= \frac{2 π}{∣ 1 ∣}

Simplify

= 2 π

Combine periods:

2 π, 2 π

= 2 π

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

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

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (x) sin (x) = sin^{1 + 1} (x)

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

Least Common Multiplier of

cos (x), 2 cos (x) : 2 cos (x)

cos (x), 2 cos (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (x)

2 cos (x)

= 2 cos (x)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

2 cos (x)

For

\frac{sin ^{2} ( x )}{cos ( x )} :

multiply the denominator and numerator by

2

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Find the zeroes and undifined points of

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

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

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

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

2 sin^{2} (x) - 1 = 0

Solve by substitution

2 sin^{2} (x) - 1 = 0

Let:

sin (x) = u

2 u^{2} - 1 = 0

2 u^{2} - 1 = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

2 u^{2} - 1 = 0

Move

1

to the right side

2 u^{2} - 1 = 0

Add

1

to both sides

2 u^{2} - 1 + 1 = 0 + 1

Simplify

2 u^{2} = 1

2 u^{2} = 1

Divide both sides by

2

2 u^{2} = 1

Divide both sides by

2

\frac{2 u ^{2}}{2} = \frac{1}{2}

Simplify

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1}{2}, u = - \frac{1}{2}

Substitute back

u = sin (x)

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

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

sin (x) = \frac{1}{2}, 0 \leq x < 2 π : x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = \frac{1}{2}, 0 \leq x < 2 π

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}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

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

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = - \frac{1}{2}, 0 \leq x < 2 π : x = \frac{5 π}{4}, x = \frac{7 π}{4}

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

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{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{5 π}{4}, x = \frac{7 π}{4}

Combine all the solutions

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

Find the undefined points:

x = \frac{π}{2}, x = \frac{3 π}{2}

Find the zeros of the denominator

2 cos (x) = 0

Divide both sides by

2

2 cos (x) = 0

Divide both sides by

2

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

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}{4}, \frac{π}{2}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}

Identify the intervals

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

Summarize in a table:

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

Identify the intervals that satisfy the required condition:

> 0

\frac{π}{4} < x < \frac{π}{2} or \frac{3 π}{4} < x < \frac{5 π}{4} or \frac{3 π}{2} < x < \frac{7 π}{4}

Apply the periodicity of

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

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

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