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Popular Trigonometry >

sin(x)+cos^2(x)<1

Solution

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

Solution

- π + 2 πn < x < 2 πn

Interval Notation

(- π + 2 πn, 2 πn)

Decimal

- 3.14159 \dots + 2 πn < x < 2 πn

Solution steps

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

Use the following identity:

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

Therefore

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

sin (x) + 1 - sin^{2} (x) < 1

Let:

u = sin (x)

u + 1 - u^{2} < 1

u + 1 - u^{2} < 1 : u < 0 or u > 1

u + 1 - u^{2} < 1

Rewrite in standard form

u + 1 - u^{2} < 1

Subtract

1

from both sides

u + 1 - u^{2} - 1 < 1 - 1

Simplify

- u^{2} + u < 0

- u^{2} + u < 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) < 0

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

u (u - 1) > 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:

> 0

u < 0 or u > 1

u < 0 or u > 1

u < 0 or u > 1

Substitute back

u = sin (x)

sin (x) < 0 or sin (x) > 1

sin (x) < 0 : - π + 2 πn < x < 2 πn

sin (x) < 0

For

sin (x) < a

, if

- 1 < a \leq 1

then

- π - arcsin (a) + 2 πn < x < arcsin (a) + 2 πn

- π - arcsin (0) + 2 πn < x < arcsin (0) + 2 πn

Simplify

- π - arcsin (0) : - π

- π - arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

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 - 0

- π - 0 = - π

= - π

Simplify

arcsin (0) : 0

arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

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}

= 0

- π + 2 πn < x < 0 + 2 πn

Simplify

- π + 2 πn < x < 2 πn

sin (x) > 1 :

False for all

x \in R

sin (x) > 1

Range of

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

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (x) \leq 1

- 1 \leq sin (x) \leq 1

sin (x) > 1 and - 1 \leq sin (x) \leq 1 :

False

Let

y = sin (x)

Combine the intervals

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

Merge Overlapping Intervals

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

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

y > 1

and

- 1 \leq y \leq 1

False for all y \in R

False for all y \in R

No Solution for x \in R

False for all x \in R

Combine the intervals

- π + 2 πn < x < 2 πn or False for all x \in R

Merge Overlapping Intervals

- π + 2 πn < x < 2 πn

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