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Popular >

2sin^2(x)+3sin(x)>= 2

Solution

2 sin^{2} (x) + 3 sin (x) \geq 2

Solution

\frac{π}{6} + 2 πn \leq x \leq \frac{5 π}{6} + 2 πn

Interval Notation

[\frac{π}{6} + 2 πn, \frac{5 π}{6} + 2 πn]

Decimal Notation

0.52359 \dots + 2 πn \leq x \leq 2.61799 \dots + 2 πn

Solution steps

2 sin^{2} (x) + 3 sin (x) \geq 2

Let:

u = sin (x)

2 u^{2} + 3 u \geq 2

2 u^{2} + 3 u \geq 2 : u \leq - 2 or u \geq \frac{1}{2}

2 u^{2} + 3 u \geq 2

Rewrite in standard form

2 u^{2} + 3 u \geq 2

Subtract

2

from both sides

2 u^{2} + 3 u - 2 \geq 2 - 2

Simplify

2 u^{2} + 3 u - 2 \geq 0

2 u^{2} + 3 u - 2 \geq 0

Factor

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

2 u^{2} + 3 u - 2

Break the expression into groups

2 u^{2} + 3 u - 2

Definition

Factors of

4 : 1, 2, 4

4

Divisors (Factors)

Find the Prime factors of

4 : 2, 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2

Add the prime factors:

2

Add 1 and the number

4

itself

1, 4

The factors of

4

1, 2, 4

Negative factors of

4 : - 1, - 2, - 4

Multiply the factors by

- 1

to get the negative factors

- 1, - 2, - 4

For every two factors such that

u * v = - 4,

check if

u + v = 3

Check

u = 1, v = - 4 : u * v = - 4, u + v = - 3 \Rightarrow

FalseCheck

u = 2, v = - 2 : u * v = - 4, u + v = 0 \Rightarrow

False

u = 4, v = - 1

Group into

(a x^{2} + ux) + (vx + c)

(2 u^{2} - u) + (4 u - 2)

= (2 u^{2} - u) + (4 u - 2)

Factor out

u

from

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

2 u^{2} - u

Apply exponent rule:

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

u^{2} = uu

= 2 uu - u

Factor out common term

u

= u (2 u - 1)

Factor out

2

from

4 u - 2 : 2 (2 u - 1)

4 u - 2

Rewrite

4

2 \cdot 2

= 2 \cdot 2 u - 2

Factor out common term

2

= 2 (2 u - 1)

= u (2 u - 1) + 2 (2 u - 1)

Factor out common term

2 u - 1

= (2 u - 1) (u + 2)

(2 u - 1) (u + 2) \geq 0

Identify the intervals

Find the signs of the factors of

(2 u - 1) (u + 2)

Find the signs of

2 u - 1

2 u - 1 = 0 : u = \frac{1}{2}

2 u - 1 = 0

Move

1

to the right side

2 u - 1 = 0

Add

1

to both sides

2 u - 1 + 1 = 0 + 1

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

2 u - 1 < 0 : u < \frac{1}{2}

2 u - 1 < 0

Move

1

to the right side

2 u - 1 < 0

Add

1

to both sides

2 u - 1 + 1 < 0 + 1

Simplify

2 u < 1

2 u < 1

Divide both sides by

2

2 u < 1

Divide both sides by

2

\frac{2 u}{2} < \frac{1}{2}

Simplify

u < \frac{1}{2}

u < \frac{1}{2}

2 u - 1 > 0 : u > \frac{1}{2}

2 u - 1 > 0

Move

1

to the right side

2 u - 1 > 0

Add

1

to both sides

2 u - 1 + 1 > 0 + 1

Simplify

2 u > 1

2 u > 1

Divide both sides by

2

2 u > 1

Divide both sides by

2

\frac{2 u}{2} > \frac{1}{2}

Simplify

u > \frac{1}{2}

u > \frac{1}{2}

Find the signs of

u + 2

u + 2 = 0 : u = - 2

u + 2 = 0

Move

2

to the right side

u + 2 = 0

Subtract

2

from both sides

u + 2 - 2 = 0 - 2

Simplify

u = - 2

u = - 2

u + 2 < 0 : u < - 2

u + 2 < 0

Move

2

to the right side

u + 2 < 0

Subtract

2

from both sides

u + 2 - 2 < 0 - 2

Simplify

u < - 2

u < - 2

u + 2 > 0 : u > - 2

u + 2 > 0

Move

2

to the right side

u + 2 > 0

Subtract

2

from both sides

u + 2 - 2 > 0 - 2

Simplify

u > - 2

u > - 2

Summarize in a table:

2 u - 1 u + 2 (2 u - 1) (u + 2) u < - 2 - - + u = - 2 - 00 - 2 < u < \frac{1}{2} - + - u = \frac{1}{2} 0 + 0 u > \frac{1}{2} + + +

Identify the intervals that satisfy the required condition:

\geq 0

u < - 2 or u = - 2 or u = \frac{1}{2} or u > \frac{1}{2}

Merge Overlapping Intervals

u \leq - 2 or u = \frac{1}{2} or u > \frac{1}{2}

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

u < - 2

u = - 2

u \leq - 2

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

u \leq - 2

u = \frac{1}{2}

u \leq - 2 or u = \frac{1}{2}

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

u \leq - 2 or u = \frac{1}{2}

u > \frac{1}{2}

u \leq - 2 or u \geq \frac{1}{2}

u \leq - 2 or u \geq \frac{1}{2}

u \leq - 2 or u \geq \frac{1}{2}

u \leq - 2 or u \geq \frac{1}{2}

Substitute back

u = sin (x)

sin (x) \leq - 2 or sin (x) \geq \frac{1}{2}

sin (x) \leq - 2 :

False for all

x \in R

sin (x) \leq - 2

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) \leq - 2 and - 1 \leq sin (x) \leq 1 :

False

Let

y = sin (x)

Combine the intervals

y \leq - 2 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \leq - 2 and - 1 \leq y \leq 1

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

y \leq - 2

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

sin (x) \geq \frac{1}{2} : \frac{π}{6} + 2 πn \leq x \leq \frac{5 π}{6} + 2 πn

sin (x) \geq \frac{1}{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{1}{2}) + 2 πn \leq x \leq π - arcsin (\frac{1}{2}) + 2 πn

Simplify

arcsin (\frac{1}{2}) : \frac{π}{6}

arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

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}{6}

Simplify

π - arcsin (\frac{1}{2}) : \frac{5 π}{6}

π - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

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}{6}

Simplify

π - \frac{π}{6}

Convert element to fraction:

π = \frac{π 6}{6}

= \frac{π 6}{6} - \frac{π}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{π 6 - π}{6}

Add similar elements:

6 π - π = 5 π

= \frac{5 π}{6}

= \frac{5 π}{6}

\frac{π}{6} + 2 πn \leq x \leq \frac{5 π}{6} + 2 πn

Combine the intervals

False for all x \in R or \frac{π}{6} + 2 πn \leq x \leq \frac{5 π}{6} + 2 πn

Merge Overlapping Intervals

\frac{π}{6} + 2 πn \leq x \leq \frac{5 π}{6} + 2 πn

Popular Examples

cos(x)<=-1/2

cos (x) \leq - \frac{1}{2}

2cos^2(2x)-3cos(2x)+1<0

2 cos^{2} (2 x) - 3 cos (2 x) + 1 < 0

cos^2(x)<=-sin(x)

cos^{2} (x) \leq - sin (x)

tan(θ)>0,sin(θ)<0

tan (θ) > 0, sin (θ) < 0

sin^2(x)> 15/40

sin^{2} (x) > \frac{15}{40}