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Popular Trigonometry >

sin(x)+cos(x)>= 1

Solution

sin (x) + cos (x) \geq 1

Solution

2 πn \leq x \leq \frac{π}{2} + 2 πn

Interval Notation

[2 πn, \frac{π}{2} + 2 πn]

Decimal

2 πn \leq x \leq 1.57079 \dots + 2 πn

Solution steps

sin (x) + cos (x) \geq 1

Use the following identity:

cos (x) + sin (x) = 2 sin (\frac{π}{4} + x)

2 sin (\frac{π}{4} + x) \geq 1

Divide both sides by

2

2 sin (\frac{π}{4} + x) \geq 1

Divide both sides by

2

\frac{2 sin ( \frac{π}{4} + x )}{2} \geq \frac{1}{2}

Simplify

\frac{2 sin ( \frac{π}{4} + x )}{2} \geq \frac{1}{2}

Simplify

\frac{2 sin ( \frac{π}{4} + x )}{2} : sin (\frac{π}{4} + x)

\frac{2 sin ( \frac{π}{4} + x )}{2}

Cancel the common factor:

2

= sin (\frac{π}{4} + x)

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

sin (\frac{π}{4} + x) \geq \frac{2}{2}

sin (\frac{π}{4} + x) \geq \frac{2}{2}

sin (\frac{π}{4} + x) \geq \frac{2}{2}

For

sin (x) \geq a

, if

- 1 < a < 1

then

arcsin (a) + 2 πn \leq x \leq π - arcsin (a) + 2 πn

arcsin (\frac{2}{2}) + 2 πn \leq (\frac{π}{4} + x) \leq π - arcsin (\frac{2}{2}) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

arcsin (\frac{2}{2}) + 2 πn \leq \frac{π}{4} + x and \frac{π}{4} + x \leq π - arcsin (\frac{2}{2}) + 2 πn

arcsin (\frac{2}{2}) + 2 πn \leq \frac{π}{4} + x : x \geq 2 πn

arcsin (\frac{2}{2}) + 2 πn \leq \frac{π}{4} + x

Switch sides

\frac{π}{4} + x \geq arcsin (\frac{2}{2}) + 2 πn

Simplify

arcsin (\frac{2}{2}) + 2 πn : \frac{π}{4} + 2 πn

arcsin (\frac{2}{2}) + 2 πn

Use the following trivial identity:

arcsin (\frac{2}{2}) = \frac{π}{4}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{4} + 2 πn

\frac{π}{4} + x \geq \frac{π}{4} + 2 πn

Move

\frac{π}{4}

to the right side

\frac{π}{4} + x \geq \frac{π}{4} + 2 πn

Subtract

\frac{π}{4}

from both sides

\frac{π}{4} + x - \frac{π}{4} \geq \frac{π}{4} + 2 πn - \frac{π}{4}

Simplify

\frac{π}{4} + x - \frac{π}{4} \geq \frac{π}{4} + 2 πn - \frac{π}{4}

Simplify

\frac{π}{4} + x - \frac{π}{4} : x

\frac{π}{4} + x - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} \geq 0

= x

Simplify

\frac{π}{4} + 2 πn - \frac{π}{4} : 2 πn

\frac{π}{4} + 2 πn - \frac{π}{4}

Group like terms

= \frac{π}{4} - \frac{π}{4} + 2 πn

Add similar elements:

\frac{π}{4} - \frac{π}{4} = 0

= 2 πn

x \geq 2 πn

x \geq 2 πn

x \geq 2 πn

\frac{π}{4} + x \leq π - arcsin (\frac{2}{2}) + 2 πn : x \leq \frac{π}{2} + 2 πn

\frac{π}{4} + x \leq π - arcsin (\frac{2}{2}) + 2 πn

Simplify

π - arcsin (\frac{2}{2}) + 2 πn : π - \frac{π}{4} + 2 πn

π - arcsin (\frac{2}{2}) + 2 πn

Use the following trivial identity:

arcsin (\frac{2}{2}) = \frac{π}{4}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= π - \frac{π}{4} + 2 πn

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

Move

\frac{π}{4}

to the right side

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

Subtract

\frac{π}{4}

from both sides

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

Simplify

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

Simplify

\frac{π}{4} + x - \frac{π}{4} : x

\frac{π}{4} + x - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} \leq 0

= x

Simplify

π - \frac{π}{4} + 2 πn - \frac{π}{4} : - \frac{π}{2} + π + 2 πn

π - \frac{π}{4} + 2 πn - \frac{π}{4}

Group like terms

= - \frac{π}{4} - \frac{π}{4} + π + 2 πn

Combine the fractions

- \frac{π}{4} - \frac{π}{4} : - \frac{π}{2}

Apply rule

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

= \frac{- π - π}{4}

Add similar elements:

- π - π = - 2 π

= \frac{- 2 π}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 π}{4}

Cancel the common factor:

2

= - \frac{π}{2}

= - \frac{π}{2} + π + 2 πn

x \leq - \frac{π}{2} + π + 2 πn

x \leq - \frac{π}{2} + π + 2 πn

x \leq - \frac{π}{2} + π + 2 πn

Simplify

- \frac{π}{2} + π : \frac{π}{2}

- \frac{π}{2} + π

Convert element to fraction:

π = \frac{π 2}{2}

= - \frac{π}{2} + \frac{π 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{- π + π 2}{2}

Add similar elements:

- π + 2 π = π

= \frac{π}{2}

x \leq \frac{π}{2} + 2 πn

Combine the intervals

x \geq 2 πn and x \leq \frac{π}{2} + 2 πn

Merge Overlapping Intervals

2 πn \leq x \leq \frac{π}{2} + 2 πn

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