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Popular >

(sin(x))/(cos(x))>= sin^2(x)+cos^2(x)

Solution

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

Solution

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

Interval Notation

[\frac{π}{4} + πn, \frac{π}{2} + πn)

Decimal Notation

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

Solution steps

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

Subtract

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

from both sides

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

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

Use the following identity:

cos^{2} (x) + sin^{2} (x) = 1

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

\frac{- cos ( x ) + sin ( x )}{cos ( x )} \geq 0

Periodicity of

- 1 + \frac{sin ( x )}{cos ( x )} : π

Periodicity of

f (x) + c =

Periodicity of

f (x)

c = - 1

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

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

is composed of the following functions and periods:

sin (x)

with periodicity of

2 π

The compound periodicity is:

= π

Find the zeroes and undifined points of

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

for

0 \leq x < π

To find the zeroes, set the inequality to zero

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

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

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

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

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

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

Add

1

to both sides

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

Find the undefined points:

x = \frac{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 < π

x = \frac{π}{2}

\frac{π}{4}, \frac{π}{2}

Identify the intervals

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

Summarize in a table:

- cos (x) + sin (x) cos (x) \frac{- c o s ( x ) + s i n ( x )}{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 < π + - - x = π + - -

Identify the intervals that satisfy the required condition:

\geq 0

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

Merge Overlapping Intervals

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

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

x = \frac{π}{4}

\frac{π}{4} < x < \frac{π}{2}

\frac{π}{4} \leq x < \frac{π}{2}

\frac{π}{4} \leq x < \frac{π}{2}

Apply the periodicity of

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

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

Popular Examples

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