English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5sin^2(t)cos(t)>0

Solution

5 sin^{2} (t) cos (t) > 0

Solution

2 πn < t < \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < t < 2 π + 2 πn

Interval Notation

(2 πn, \frac{π}{2} + 2 πn) \cup (\frac{3 π}{2} + 2 πn, 2 π + 2 πn)

Decimal Notation

2 πn < t < 1.57079 \dots + 2 πn or 4.71238 \dots + 2 πn < t < 6.28318 \dots + 2 πn

Solution steps

5 sin^{2} (t) cos (t) > 0

Use the following identity:

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

Therefore

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

5 (1 - cos^{2} (t)) cos (t) > 0

Let:

u = cos (t)

5 (1 - u^{2}) u > 0

5 (1 - u^{2}) u > 0 : u < - 1 or 0 < u < 1

5 (1 - u^{2}) u > 0

Rewrite in standard form

5 (1 - u^{2}) u > 0

Expand

5 (1 - u^{2}) u : 5 u - 5 u^{3}

5 (1 - u^{2}) u

= 5 u (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 5 u, b = 1, c = u^{2}

= 5 u \cdot 1 - 5 u u^{2}

= 5 \cdot 1 \cdot u - 5 u^{2} u

Simplify

5 \cdot 1 \cdot u - 5 u^{2} u : 5 u - 5 u^{3}

5 \cdot 1 \cdot u - 5 u^{2} u

5 \cdot 1 \cdot u = 5 u

5 \cdot 1 \cdot u

Multiply the numbers:

5 \cdot 1 = 5

= 5 u

5 u^{2} u = 5 u^{3}

5 u^{2} u

Apply exponent rule:

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

u^{2} u = u^{2 + 1}

= 5 u^{2 + 1}

Add the numbers:

2 + 1 = 3

= 5 u^{3}

= 5 u - 5 u^{3}

= 5 u - 5 u^{3}

5 u - 5 u^{3} > 0

Divide both sides by

5

\frac{5 u}{5} - \frac{5 u ^{3}}{5} > \frac{0}{5}

Refine

\frac{5 u}{5} - \frac{5 u ^{3}}{5} > \frac{0}{5} : - u^{3} + u > 0

\frac{5 u}{5} - \frac{5 u ^{3}}{5} > \frac{0}{5}

Simplify

\frac{5 u}{5} - \frac{5 u ^{3}}{5} : - u^{3} + u

\frac{5 u}{5} - \frac{5 u ^{3}}{5}

Divide the numbers:

\frac{5}{5} = 1

= u - u^{3}

Rewrite in standard form

= - u^{3} + u

\frac{0}{5} = 0

\frac{0}{5}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

- u^{3} + u > 0

- u^{3} + u > 0

- u^{3} + u > 0

Factor

- u^{3} + u : - u (u + 1) (u - 1)

- u^{3} + u

Factor out common term

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

- u^{3} + u

Apply exponent rule:

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

u^{3} = u^{2} u

= - u^{2} u + u

Factor out common term

- u

= - u (u^{2} - 1)

= - u (u^{2} - 1)

Factor

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

u^{2} - 1

Rewrite

1

1^{2}

= u^{2} - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (u + 1) (u - 1)

= - u (u + 1) (u - 1)

- u (u + 1) (u - 1) > 0

Multiply both sides by

- 1

(reverse the inequality)

(- u (u + 1) (u - 1)) (- 1) < 0 \cdot (- 1)

Simplify

u (u + 1) (u - 1) < 0

Identify the intervals

Find the signs of the factors of

u (u + 1) (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

Subtract

1

from 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

Subtract

1

from 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

Subtract

1

from both sides

u + 1 - 1 > 0 - 1

Simplify

u > - 1

u > - 1

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 - 1 u (u + 1) (u - 1) u < - 1 - - - - u = - 1 - 0 - 0 - 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 < - 1 or 0 < u < 1

u < - 1 or 0 < u < 1

u < - 1 or 0 < u < 1

Substitute back

u = cos (t)

cos (t) < - 1 or 0 < cos (t) < 1

cos (t) < - 1 :

False for all

t \in R

cos (t) < - 1

Range of

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

Function range definition

The range of the basic

cos

function is

- 1 \leq cos (t) \leq 1

- 1 \leq cos (t) \leq 1

cos (t) < - 1 and - 1 \leq cos (t) \leq 1 :

False

Let

y = cos (t)

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 t \in R

False for all t \in R

0 < cos (t) < 1 : 2 πn < t < \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < t < 2 π + 2 πn

0 < cos (t) < 1

a < u < b

then

a < u and u < b

0 < cos (t) and cos (t) < 1

0 < cos (t) : - \frac{π}{2} + 2 πn < t < \frac{π}{2} + 2 πn

0 < cos (t)

Switch sides

cos (t) > 0

For

cos (x) > a

, if

- 1 \leq a < 1

then

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (0) + 2 πn < t < 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 < t < \frac{π}{2} + 2 πn

cos (t) < 1 : 2 πn < t < 2 π + 2 πn

cos (t) < 1

For

cos (x) < a

, if

- 1 < a \leq 1

then

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

arccos (1) + 2 πn < t < 2 π - arccos (1) + 2 πn

Simplify

arccos (1) : 0

arccos (1)

Use the following trivial identity:

arccos (1) = 0

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}

= 0

Simplify

2 π - arccos (1) : 2 π

2 π - arccos (1)

Use the following trivial identity:

arccos (1) = 0

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}

= 2 π - 0

2 π - 0 = 2 π

= 2 π

0 + 2 πn < t < 2 π + 2 πn

Simplify

2 πn < t < 2 π + 2 πn

Combine the intervals

- \frac{π}{2} + 2 πn < t < \frac{π}{2} + 2 πn and 2 πn < t < 2 π + 2 πn

Merge Overlapping Intervals

2 πn < t < \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < t < 2 π + 2 πn

Combine the intervals

False for all t \in R or (2 πn < t < \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < t < 2 π + 2 πn)

Merge Overlapping Intervals

2 πn < t < \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < t < 2 π + 2 πn

Popular Examples

(2cos(2x)+1)(tan(x)-1)<= 0

(2 cos (2 x) + 1) (tan (x) - 1) \leq 0

cos(x)+sin(x)>0

cos (x) + sin (x) > 0

sqrt(2)cos(2x)<1

2 cos (2 x) < 1

sec^3(x)+sec(x)tan^2(x)>0

sec^{3} (x) + sec (x) tan^{2} (x) > 0

12cos(2x)<0

12 cos (2 x) < 0