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Popular Trigonometry >

cos^2(x)>sin(x)cos(x)

Solution

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

Solution

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

Interval Notation

[πn, \frac{π}{4} + πn) \cup (\frac{π}{2} + πn, π + πn]

Decimal

πn \leq x < 0.78539 \dots + πn or 1.57079 \dots + πn < x \leq 3.14159 \dots + πn

Solution steps

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

Move

sin (x) cos (x)

to the left side

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

Subtract

sin (x) cos (x)

from both sides

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

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

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

Periodicity of

cos^{2} (x) - sin (x) cos (x) : π

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

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

Periodicity of

cos^{2} (x) : π

Periodicity of

cos^{n} (x) = \frac{Periodicity of cos ( x )}{2},

if n is even

Periodicity of

cos (x) : 2 π

Periodicity of

cos (x)

2 π

= 2 π

\frac{2 π}{2}

Simplify

π

Periodicity of

sin (x) cos (x) : π

sin (x) cos (x)

is composed of the following functions and periods:

cos (x)

with periodicity of

2 π

The compound periodicity is:

π

Combine periods:

π, π

= π

Factor

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

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

Apply exponent rule:

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

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

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

Factor out common term

cos (x)

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

cos (x) (cos (x) - sin (x)) > 0

To find the zeroes, set the inequality to zero

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

Solve

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

for

0 \leq x < π

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

Solving each part separately

cos (x) = 0 : x = \frac{π}{2}

cos (x) = 0, 0 \leq x < π

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 < π

x = \frac{π}{2}

cos (x) - sin (x) = 0 : x = \frac{π}{4}

cos (x) - sin (x) = 0, 0 \leq x < 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 < π

x = \frac{π}{4}

Combine all the solutions

\frac{π}{4} or \frac{π}{2}

The intervals between the zeros

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

Summarize in a table:

cos (x) cos (x) - sin (x) cos (x) (cos (x) - sin (x)) x = 0 + + + 0 < x < \frac{π}{4} + + + x = \frac{π}{4} + 00 \frac{π}{4} < x < \frac{π}{2} + - - x = \frac{π}{2} 0 - 0 \frac{π}{2} < x < π - - + x = π - - +

Identify the intervals that satisfy the required condition:

> 0

x = 0 or 0 < x < \frac{π}{4} or \frac{π}{2} < x < π or x = π

Merge Overlapping Intervals

0 \leq x < \frac{π}{4} or \frac{π}{2} < x < π or x = π

The union of two intervals is the set of numbers which are in either interval

x = 0

0 < x < \frac{π}{4}

0 \leq x < \frac{π}{4}

The union of two intervals is the set of numbers which are in either interval

0 \leq x < \frac{π}{4}

\frac{π}{2} < x < π

0 \leq x < \frac{π}{4} or \frac{π}{2} < x < π

The union of two intervals is the set of numbers which are in either interval

0 \leq x < \frac{π}{4} or \frac{π}{2} < x < π

x = π

0 \leq x < \frac{π}{4} or \frac{π}{2} < x \leq π

0 \leq x < \frac{π}{4} or \frac{π}{2} < x \leq π

Apply the periodicity of

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

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

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