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Popular Trigonometry >

2sin^2(x)-5sin(x)-3>= 0

Solution

2 sin^{2} (x) - 5 sin (x) - 3 \geq 0

Solution

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

Interval Notation

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

Decimal

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

Solution steps

2 sin^{2} (x) - 5 sin (x) - 3 \geq 0

Let:

u = sin (x)

2 u^{2} - 5 u - 3 \geq 0

2 u^{2} - 5 u - 3 \geq 0 : u \leq - \frac{1}{2} or u \geq 3

2 u^{2} - 5 u - 3 \geq 0

Factor

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

2 u^{2} - 5 u - 3

Break the expression into groups

2 u^{2} - 5 u - 3

Definition

Factors of

6 : 1, 2, 3, 6

6

Divisors (Factors)

Find the Prime factors of

6 : 2, 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Add the prime factors:

2, 3

Add 1 and the number

6

itself

1, 6

The factors of

6

1, 2, 3, 6

Negative factors of

6 : - 1, - 2, - 3, - 6

Multiply the factors by

- 1

to get the negative factors

- 1, - 2, - 3, - 6

For every two factors such that

u * v = - 6,

check if

u + v = - 5

Check

u = 1, v = - 6 : u * v = - 6, u + v = - 5 \Rightarrow

TrueCheck

u = 2, v = - 3 : u * v = - 6, u + v = - 1 \Rightarrow

False

u = 1, v = - 6

Group into

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

(2 u^{2} + u) + (- 6 u - 3)

= (2 u^{2} + u) + (- 6 u - 3)

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

- 3

from

- 6 u - 3 : - 3 (2 u + 1)

- 6 u - 3

Rewrite

6

3 \cdot 2

= - 3 \cdot 2 u - 3

Factor out common term

- 3

= - 3 (2 u + 1)

= u (2 u + 1) - 3 (2 u + 1)

Factor out common term

2 u + 1

= (2 u + 1) (u - 3)

(2 u + 1) (u - 3) \geq 0

Identify the intervals

Find the signs of the factors of

(2 u + 1) (u - 3)

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

Subtract

1

from 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

Subtract

1

from 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

Subtract

1

from 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 - 3

u - 3 = 0 : u = 3

u - 3 = 0

Move

3

to the right side

u - 3 = 0

Add

3

to both sides

u - 3 + 3 = 0 + 3

Simplify

u = 3

u = 3

u - 3 < 0 : u < 3

u - 3 < 0

Move

3

to the right side

u - 3 < 0

Add

3

to both sides

u - 3 + 3 < 0 + 3

Simplify

u < 3

u < 3

u - 3 > 0 : u > 3

u - 3 > 0

Move

3

to the right side

u - 3 > 0

Add

3

to both sides

u - 3 + 3 > 0 + 3

Simplify

u > 3

u > 3

Summarize in a table:

2 u + 1 u - 3 (2 u + 1) (u - 3) u < - \frac{1}{2} - - + u = - \frac{1}{2} 0 - 0 - \frac{1}{2} < u < 3 + - - u = 3 + 00 u > 3 + + +

Identify the intervals that satisfy the required condition:

\geq 0

u < - \frac{1}{2} or u = - \frac{1}{2} or u = 3 or u > 3

Merge Overlapping Intervals

u \leq - \frac{1}{2} or u = 3 or u > 3

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

u < - \frac{1}{2}

u = - \frac{1}{2}

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

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

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

u = 3

u \leq - \frac{1}{2} or u = 3

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

u \leq - \frac{1}{2} or u = 3

u > 3

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

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

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

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

Substitute back

u = sin (x)

sin (x) \leq - \frac{1}{2} or sin (x) \geq 3

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

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

For

sin (x) \leq a

, if

- 1 < a < 1

then

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

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

Simplify

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

- π - arcsin (- \frac{1}{2})

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

arcsin (- \frac{1}{2})

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

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}

= - \frac{π}{6}

= - π - (- \frac{π}{6})

Simplify

- π - (- \frac{π}{6})

Apply rule

- (- a) = a

= - 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}

Apply the fraction rule:

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

= - \frac{5 π}{6}

= - \frac{5 π}{6}

Simplify

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

arcsin (- \frac{1}{2})

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

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}

= - \frac{π}{6}

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

sin (x) \geq 3 :

False for all

x \in R

sin (x) \geq 3

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) \geq 3 and - 1 \leq sin (x) \leq 1 :

False

Let

y = sin (x)

Combine the intervals

y \geq 3 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \geq 3 and - 1 \leq y \leq 1

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

y \geq 3

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

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

Merge Overlapping Intervals

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

Popular Examples