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Popular Trigonometry >

2(cos(3x))^2+sqrt(3)sin(6x)<1

Solution

2 (cos (3 x))^{2} + 3 sin (6 x) < 1

Solution

\frac{π}{6} + \frac{π}{3} n < x < \frac{5 π}{18} + \frac{π}{3} n

Interval Notation

(\frac{π}{6} + \frac{π}{3} n, \frac{5 π}{18} + \frac{π}{3} n)

Decimal

0.52359 \dots + \frac{π}{3} n < x < 0.87266 \dots + \frac{π}{3} n

Solution steps

2 (cos (3 x))^{2} + 3 sin (6 x) < 1

Let:

u = 3 x

2 (cos (u))^{2} + 3 sin (2 u) < 1

2 (cos (u))^{2} + 3 sin (2 u) < 1 : \frac{π}{2} + πn < u < \frac{5 π}{6} + πn

2 (cos (u))^{2} + 3 sin (2 u) < 1

Use the following identity:

sin (2 x) = 2 cos (x) sin (x)

2 (cos (u))^{2} + 3 \cdot 2 cos (u) sin (u) < 1

Simplify

2 cos^{2} (u) + 23 cos (u) sin (u) < 1

Periodicity of

2 cos^{2} (u) + 23 cos (u) sin (u) : π

The compound periodicity of the sum of periodic functions is the least common multiplier of the periods

2 cos^{2} (u), 23 cos (u) sin (u)

Periodicity of

2 cos^{2} (u) : π

Periodicity of

cos^{n} (x) = \frac{Periodicity of cos ( x )}{2},

if n is even

Periodicity of

cos (u) : 2 π

Periodicity of

cos (x)

2 π

= 2 π

\frac{2 π}{2}

Simplify

π

Periodicity of

23 cos (u) sin (u) : π

23 cos (u) sin (u)

is composed of the following functions and periods:

cos (u)

with periodicity of

2 π

The compound periodicity is:

π

Combine periods:

π, π

= π

Factor

2 cos^{2} (u) + 23 cos (u) sin (u) : 2 cos (u) (cos (u) + 3 sin (u))

2 cos^{2} (u) + 23 cos (u) sin (u)

Apply exponent rule:

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

cos^{2} (u) = cos (u) cos (u)

= 2 cos (u) cos (u) + 23 cos (u) sin (u)

Factor out common term

2 cos (u)

= 2 cos (u) (cos (u) + 3 sin (u))

2 cos (u) (cos (u) + 3 sin (u)) < 1

To find the zeroes, set the inequality to zero

2 cos (u) (cos (u) + 3 sin (u)) = 0

Solve

2 cos (u) (cos (u) + 3 sin (u)) = 0

for

0 \leq u < π

2 cos (u) (cos (u) + 3 sin (u)) = 0

Solving each part separately

cos (u) = 0 : u = \frac{π}{2}

cos (u) = 0, 0 \leq u < π

General solutions for

cos (u) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq u < π

u = \frac{π}{2}

cos (u) + 3 sin (u) = 0 : u = \frac{5 π}{6}

cos (u) + 3 sin (u) = 0, 0 \leq u < π

Rewrite using trig identities

cos (u) + 3 sin (u) = 0

Divide both sides by

cos (u), cos (u) \neq = 0

\frac{cos ( u ) + 3 sin ( u )}{cos ( u )} = \frac{0}{cos ( u )}

Simplify

1 + \frac{3 sin ( u )}{cos ( u )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 + 3 tan (u) = 0

1 + 3 tan (u) = 0

Move

1

to the right side

1 + 3 tan (u) = 0

Subtract

1

from both sides

1 + 3 tan (u) - 1 = 0 - 1

Simplify

3 tan (u) = - 1

3 tan (u) = - 1

Divide both sides by

3

3 tan (u) = - 1

Divide both sides by

3

\frac{3 tan ( u )}{3} = \frac{- 1}{3}

Simplify

\frac{3 tan ( u )}{3} = \frac{- 1}{3}

Simplify

\frac{3 tan ( u )}{3} : tan (u)

\frac{3 tan ( u )}{3}

Cancel the common factor:

3

= tan (u)

Simplify

\frac{- 1}{3} : - \frac{3}{3}

\frac{- 1}{3}

Apply the fraction rule:

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

= - \frac{1}{3}

Rationalize

- \frac{1}{3} : - \frac{3}{3}

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= - \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

tan (u) = - \frac{3}{3}

tan (u) = - \frac{3}{3}

tan (u) = - \frac{3}{3}

General solutions for

tan (u) = - \frac{3}{3}

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

u = \frac{5 π}{6} + πn

u = \frac{5 π}{6} + πn

Solutions for the range

0 \leq u < π

u = \frac{5 π}{6}

Combine all the solutions

\frac{π}{2} or \frac{5 π}{6}

The intervals between the zeros

0 < u < \frac{π}{2}, \frac{π}{2} < u < \frac{5 π}{6}, \frac{5 π}{6} < u < π

Summarize in a table:

cos (u) cos (u) + 3 sin (u) 2 cos (u) (cos (u) + 3 sin (u)) u = 0 + + + 0 < u < \frac{π}{2} + + + u = \frac{π}{2} 0 + 0 \frac{π}{2} < u < \frac{5 π}{6} - + - u = \frac{5 π}{6} - 00 \frac{5 π}{6} < u < π - - + u = π - - +

Identify the intervals that satisfy the required condition:

< 0

\frac{π}{2} < u < \frac{5 π}{6}

Apply the periodicity of

2 cos^{2} (u) + 23 cos (u) sin (u)

\frac{π}{2} + πn < u < \frac{5 π}{6} + πn

\frac{π}{2} + πn < u < \frac{5 π}{6} + πn

Substitute back

3 x = u

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

\frac{π}{2} + πn < 3 x < \frac{5 π}{6} + πn : \frac{π}{6} + \frac{π}{3} n < x < \frac{5 π}{18} + \frac{π}{3} n

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

a < u < b

then

a < u and u < b

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

\frac{π}{2} + πn < 3 x : x > \frac{π}{6} + \frac{πn}{3}

\frac{π}{2} + πn < 3 x

Switch sides

3 x > \frac{π}{2} + πn

Divide both sides by

3

3 x > \frac{π}{2} + πn

Divide both sides by

3

\frac{3 x}{3} > \frac{\frac{π}{2}}{3} + \frac{πn}{3}

Simplify

\frac{3 x}{3} > \frac{\frac{π}{2}}{3} + \frac{πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{π}{2}}{3} + \frac{πn}{3} : \frac{π}{6} + \frac{πn}{3}

\frac{\frac{π}{2}}{3} + \frac{πn}{3}

\frac{\frac{π}{2}}{3} = \frac{π}{6}

\frac{\frac{π}{2}}{3}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{π}{6}

= \frac{π}{6} + \frac{πn}{3}

x > \frac{π}{6} + \frac{πn}{3}

x > \frac{π}{6} + \frac{πn}{3}

x > \frac{π}{6} + \frac{πn}{3}

3 x < \frac{5 π}{6} + πn : x < \frac{5 π}{18} + \frac{πn}{3}

3 x < \frac{5 π}{6} + πn

Divide both sides by

3

3 x < \frac{5 π}{6} + πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{5 π}{6}}{3} + \frac{πn}{3} : \frac{5 π}{18} + \frac{πn}{3}

\frac{\frac{5 π}{6}}{3} + \frac{πn}{3}

\frac{\frac{5 π}{6}}{3} = \frac{5 π}{18}

\frac{\frac{5 π}{6}}{3}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{5 π}{18}

= \frac{5 π}{18} + \frac{πn}{3}

x < \frac{5 π}{18} + \frac{πn}{3}

x < \frac{5 π}{18} + \frac{πn}{3}

x < \frac{5 π}{18} + \frac{πn}{3}

Combine the intervals

x > \frac{π}{6} + \frac{π}{3} n and x < \frac{5 π}{18} + \frac{π}{3} n

Merge Overlapping Intervals

\frac{π}{6} + \frac{π}{3} n < x < \frac{5 π}{18} + \frac{π}{3} n

\frac{π}{6} + \frac{π}{3} n < x < \frac{5 π}{18} + \frac{π}{3} n

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