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Popular Trigonometry >

(tan(x)-tan^2(x))/(2sin(x)-1)<0

Solution

\frac{tan ( x ) - tan ^{2} ( x )}{2 sin ( x ) - 1} < 0

Solution

2 πn < x < \frac{π}{6} + 2 πn or \frac{π}{4} + 2 πn < x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{5 π}{6} + 2 πn or π + 2 πn < x < \frac{5 π}{4} + 2 πn

Interval Notation

(2 πn, \frac{π}{6} + 2 πn) \cup (\frac{π}{4} + 2 πn, \frac{π}{2} + 2 πn) \cup (\frac{π}{2} + 2 πn, \frac{5 π}{6} + 2 πn) \cup (π + 2 πn, \frac{5 π}{4} + 2 πn)

Decimal

2 πn < x < 0.52359 \dots + 2 πn or 0.78539 \dots + 2 πn < x < 1.57079 \dots + 2 πn or 1.57079 \dots + 2 πn < x < 2.61799 \dots + 2 πn or 3.14159 \dots + 2 πn < x < 3.92699 \dots + 2 πn

Solution steps

\frac{tan ( x ) - tan ^{2} ( x )}{2 sin ( x ) - 1} < 0

Periodicity of

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

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

is composed of the following functions and periods:

tan (x)

with periodicity of

π

The compound periodicity is:

= 2 π

Express with sin, cos

\frac{tan ( x ) - tan ^{2} ( x )}{2 sin ( x ) - 1} < 0

Use the basic trigonometric identity:

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

\frac{\frac{s i n ( x )}{c o s ( x )} - ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{2 sin ( x ) - 1} < 0

\frac{\frac{s i n ( x )}{c o s ( x )} - ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{2 sin ( x ) - 1} < 0

Simplify

\frac{\frac{s i n ( x )}{c o s ( x )} - ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{2 sin ( x ) - 1} : \frac{sin ( x ) cos ( x ) - sin ^{2} ( x )}{cos ^{2} ( x ) ( 2 sin ( x ) - 1 )}

\frac{\frac{s i n ( x )}{c o s ( x )} - ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{2 sin ( x ) - 1}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{\frac{s i n ( x )}{c o s ( x )} - \frac{s i n ^{2} ( x )}{c o s ^{2} ( x )}}{2 sin ( x ) - 1}

Join

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

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

Least Common Multiplier of

cos (x), cos^{2} (x) : cos^{2} (x)

cos (x), cos^{2} (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (x)

cos^{2} (x)

= cos^{2} (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

cos^{2} (x)

For

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

multiply the denominator and numerator by

cos (x)

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

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

Since the denominators are equal, combine the fractions:

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

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

= \frac{\frac{s i n ( x ) c o s ( x ) - s i n ^{2} ( x )}{c o s ^{2} ( x )}}{2 sin ( x ) - 1}

Apply the fraction rule:

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

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

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

Find the zeroes and undifined points of

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

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

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

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

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

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

Factor

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

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

Apply exponent rule:

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

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

= sin (x) cos (x) - sin (x) sin (x)

Factor out common term

sin (x)

= sin (x) (cos (x) - sin (x))

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

Solving each part separately

sin (x) = 0 or cos (x) - sin (x) = 0

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Solutions for the range

0 \leq x < 2 π

x = 0, x = π

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

cos (x) - sin (x) = 0, 0 \leq x < 2 π

Rewrite using trig identities

cos (x) - sin (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

1 - tan (x) = 0

1 - tan (x) = 0

Move

1

to the right side

1 - tan (x) = 0

Subtract

1

from both sides

1 - tan (x) - 1 = 0 - 1

Simplify

- tan (x) = - 1

- tan (x) = - 1

Divide both sides by

- 1

- tan (x) = - 1

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{- 1}{- 1}

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

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}

Combine all the solutions

x = 0, x = π, x = \frac{π}{4}, x = \frac{5 π}{4}

Find the undefined points:

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

Find the zeros of the denominator

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

Solving each part separately

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

cos^{2} (x) = 0, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos^{2} (x) = 0, 0 \leq x < 2 π

Apply rule

x^{n} = 0 \Rightarrow 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}

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

2 sin (x) - 1 = 0, 0 \leq x < 2 π

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}

Combine all the solutions

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

0, \frac{π}{6}, \frac{π}{4}, \frac{π}{2}, \frac{5 π}{6}, π, \frac{5 π}{4}, \frac{3 π}{2}

Identify the intervals

0 < x < \frac{π}{6}, \frac{π}{6} < x < \frac{π}{4}, \frac{π}{4} < x < \frac{π}{2}, \frac{π}{2} < x < \frac{5 π}{6}, \frac{5 π}{6} < x < π, π < 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:

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

Identify the intervals that satisfy the required condition:

< 0

0 < x < \frac{π}{6} or \frac{π}{4} < x < \frac{π}{2} or \frac{π}{2} < x < \frac{5 π}{6} or π < x < \frac{5 π}{4}

Apply the periodicity of

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

2 πn < x < \frac{π}{6} + 2 πn or \frac{π}{4} + 2 πn < x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{5 π}{6} + 2 πn or π + 2 πn < x < \frac{5 π}{4} + 2 πn

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