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Popular Trigonometry >

sin^2(x)+cos(x)>= 1

Solution

sin^{2} (x) + cos (x) \geq 1

Solution

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

Interval Notation

[- \frac{π}{2} + 2 πn, \frac{π}{2} + 2 πn]

Decimal

- 1.57079 \dots + 2 πn \leq x \leq 1.57079 \dots + 2 πn

Solution steps

sin^{2} (x) + cos (x) \geq 1

Use the following identity:

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

Therefore

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

1 - cos^{2} (x) + cos (x) \geq 1

Let:

u = cos (x)

1 - u^{2} + u \geq 1

1 - u^{2} + u \geq 1 : 0 \leq u \leq 1

1 - u^{2} + u \geq 1

Rewrite in standard form

1 - u^{2} + u \geq 1

Subtract

1

from both sides

1 - u^{2} + u - 1 \geq 1 - 1

Simplify

- u^{2} + u \geq 0

- u^{2} + u \geq 0

Factor

- u^{2} + u : - u (u - 1)

- u^{2} + u

Apply exponent rule:

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

u^{2} = uu

= - uu + u

Factor out common term

- u

= - u (u - 1)

- u (u - 1) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- u (u - 1)) (- 1) \leq 0 \cdot (- 1)

Simplify

u (u - 1) \leq 0

Identify the intervals

Find the signs of the factors of

u (u - 1)

Find the signs of

u

u = 0

u < 0

u > 0

Find the signs of

u - 1

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

u - 1 < 0 : u < 1

u - 1 < 0

Move

1

to the right side

u - 1 < 0

Add

1

to both sides

u - 1 + 1 < 0 + 1

Simplify

u < 1

u < 1

u - 1 > 0 : u > 1

u - 1 > 0

Move

1

to the right side

u - 1 > 0

Add

1

to both sides

u - 1 + 1 > 0 + 1

Simplify

u > 1

u > 1

Summarize in a table:

u u - 1 u (u - 1) u < 0 - - + u = 0 0 - 0 0 < u < 1 + - - u = 1 + 00 u > 1 + + +

Identify the intervals that satisfy the required condition:

\leq 0

u = 0 or 0 < u < 1 or u = 1

Merge Overlapping Intervals

0 \leq u < 1 or u = 1

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

u = 0

0 < u < 1

0 \leq u < 1

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

0 \leq u < 1

u = 1

0 \leq u \leq 1

0 \leq u \leq 1

0 \leq u \leq 1

0 \leq u \leq 1

Substitute back

u = cos (x)

0 \leq cos (x) \leq 1

a \leq u \leq b

then

a \leq u and u \leq b

0 \leq cos (x) and cos (x) \leq 1

0 \leq cos (x) : - \frac{π}{2} + 2 πn \leq x \leq \frac{π}{2} + 2 πn

0 \leq cos (x)

Switch sides

cos (x) \geq 0

For

cos (x) \geq a

, if

- 1 < a < 1

then

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

- arccos (0) + 2 πn \leq x \leq arccos (0) + 2 πn

Simplify

- arccos (0) : - \frac{π}{2}

- arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{2}

Simplify

arccos (0) : \frac{π}{2}

arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2}

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

cos (x) \leq 1 :

True for all

x \in R

cos (x) \leq 1

Range of

cos (x) : - 1 \leq cos (x) \leq 1

Function range definition

The range of the basic

cos

function is

- 1 \leq cos (x) \leq 1

- 1 \leq cos (x) \leq 1

cos (x) \leq 1 and - 1 \leq cos (x) \leq 1 : - 1 \leq cos (x) \leq 1

Let

y = cos (x)

Combine the intervals

y \leq 1 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \leq 1 and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y \leq 1

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all x

True for all x \in R

Combine the intervals

- \frac{π}{2} + 2 πn \leq x \leq \frac{π}{2} + 2 πn and True for all x \in R

Merge Overlapping Intervals

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

Popular Examples

cos(x)>= (sqrt(3))/2

cos (x) \geq \frac{3}{2}

cos(x)-1>= 0

cos (x) - 1 \geq 0

2sin(x/2)-1>0

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

sin(3x)<(sqrt(2))/2

sin (3 x) < \frac{2}{2}

sin(2x-pi/(12))<= (sqrt(2))/2

sin (2 x - \frac{π}{12}) \leq \frac{2}{2}