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Popular >

2(cos(x))^2+5sin(x)-3>2sin(x)

Solution

2 (cos (x))^{2} + 5 sin (x) - 3 > 2 sin (x)

Solution

\frac{π}{6} + 2 πn < x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{5 π}{6} + 2 πn

Interval Notation

(\frac{π}{6} + 2 πn, \frac{π}{2} + 2 πn) \cup (\frac{π}{2} + 2 πn, \frac{5 π}{6} + 2 πn)

Decimal Notation

0.52359 \dots + 2 πn < x < 1.57079 \dots + 2 πn or 1.57079 \dots + 2 πn < x < 2.61799 \dots + 2 πn

Solution steps

2 (cos (x))^{2} + 5 sin (x) - 3 > 2 sin (x)

Move

2 sin (x)

to the left side

2 (cos (x))^{2} + 5 sin (x) - 3 > 2 sin (x)

Subtract

2 sin (x)

from both sides

2 (cos (x))^{2} + 5 sin (x) - 3 - 2 sin (x) > 2 sin (x) - 2 sin (x)

2 (cos (x))^{2} + 5 sin (x) - 3 - 2 sin (x) > 2 sin (x) - 2 sin (x)

Refine

Simplify

2 (cos (x))^{2} + 5 sin (x) - 3 - 2 sin (x) : 2 cos^{2} (x) + 3 sin (x) - 3

2 (cos (x))^{2} + 5 sin (x) - 3 - 2 sin (x)

Group like terms

= 2 cos^{2} (x) + 5 sin (x) - 2 sin (x) - 3

Add similar elements:

5 sin (x) - 2 sin (x) = 3 sin (x)

= 2 cos^{2} (x) + 3 sin (x) - 3

2 sin (x) - 2 sin (x)

Add similar elements:

2 sin (x) - 2 sin (x) > 0

= 0

2 cos^{2} (x) + 3 sin (x) - 3 > 0

2 cos^{2} (x) + 3 sin (x) - 3 > 0

2 cos^{2} (x) + 3 sin (x) - 3 > 0

Use the following identity:

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

Therefore

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

2 (1 - sin^{2} (x)) + 3 sin (x) - 3 > 0

Simplify

2 (1 - sin^{2} (x)) + 3 sin (x) - 3 : 3 sin (x) - 2 sin^{2} (x) - 1

2 (1 - sin^{2} (x)) + 3 sin (x) - 3

Expand

2 (1 - sin^{2} (x)) : 2 - 2 sin^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (x)

= 2 \cdot 1 - 2 sin^{2} (x)

Multiply the numbers:

2 \cdot 1 = 2

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

= 2 - 2 sin^{2} (x) + 3 sin (x) - 3

Simplify

2 - 2 sin^{2} (x) + 3 sin (x) - 3 : 3 sin (x) - 2 sin^{2} (x) - 1

2 - 2 sin^{2} (x) + 3 sin (x) - 3

Group like terms

= - 2 sin^{2} (x) + 3 sin (x) + 2 - 3

Add/Subtract the numbers:

2 - 3 = - 1

= 3 sin (x) - 2 sin^{2} (x) - 1

= 3 sin (x) - 2 sin^{2} (x) - 1

3 sin (x) - 2 sin^{2} (x) - 1 > 0

Let:

u = sin (x)

3 u - 2 u^{2} - 1 > 0

3 u - 2 u^{2} - 1 > 0 : \frac{1}{2} < u < 1

3 u - 2 u^{2} - 1 > 0

Factor

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

3 u - 2 u^{2} - 1

Factor out common term

- 1

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

Factor

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

2 u^{2} - 3 u + 1

Write in the standard form

a x^{2} + b x + c

= 2 u^{2} - 3 u + 1

Break the expression into groups

2 u^{2} - 3 u + 1

Definition

Factors of

2 : 1, 2

2

Divisors (Factors)

Find the Prime factors of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Add 1

1

The factors of

2

1, 2

Negative factors of

2 : - 1, - 2

Multiply the factors by

- 1

to get the negative factors

- 1, - 2

For every two factors such that

u * v = 2,

check if

u + v = - 3

Check

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

FalseCheck

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

True

u = - 1, v = - 2

Group into

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

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

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

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

- 1

from

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

- 2 u + 1

Factor out common term

- 1

= - (2 u - 1)

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

Factor out common term

2 u - 1

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

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

- (2 u - 1) (u - 1) > 0

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

(2 u - 1) (u - 1) < 0

Identify the intervals

Find the signs of the factors of

(2 u - 1) (u - 1)

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 - 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:

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

Identify the intervals that satisfy the required condition:

< 0

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

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

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

Substitute back

u = sin (x)

\frac{1}{2} < sin (x) < 1

a < u < b

then

a < u and u < b

\frac{1}{2} < sin (x) and sin (x) < 1

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

\frac{1}{2} < sin (x)

Switch sides

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

For

sin (x) > a

, if

- 1 \leq a < 1

then

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

arcsin (\frac{1}{2}) + 2 πn < x < 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 < x < \frac{5 π}{6} + 2 πn

sin (x) < 1 : - \frac{3 π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

sin (x) < 1

For

sin (x) < a

, if

- 1 < a \leq 1

then

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

- π - arcsin (1) + 2 πn < x < arcsin (1) + 2 πn

Simplify

- π - arcsin (1) : - \frac{3 π}{2}

- π - arcsin (1)

Use the following trivial identity:

arcsin (1) = \frac{π}{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}

= - π - \frac{π}{2}

Simplify

- π - \frac{π}{2}

Convert element to fraction:

π = \frac{π 2}{2}

= - \frac{π 2}{2} - \frac{π}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 2 - π}{2}

Add similar elements:

- 2 π - π = - 3 π

= \frac{- 3 π}{2}

Apply the fraction rule:

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

= - \frac{3 π}{2}

= - \frac{3 π}{2}

Simplify

arcsin (1) : \frac{π}{2}

arcsin (1)

Use the following trivial identity:

arcsin (1) = \frac{π}{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}{2}

- \frac{3 π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

Combine the intervals

\frac{π}{6} + 2 πn < x < \frac{5 π}{6} + 2 πn and - \frac{3 π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

Merge Overlapping Intervals

\frac{π}{6} + 2 πn < x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{5 π}{6} + 2 πn

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tan (5 x) \leq 1

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