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Popular >

sin(2x)>6cos(x)

Solution

sin (2 x) > 6 cos (x)

Solution

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

Interval Notation

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

Decimal Notation

1.57079 \dots + 2 πn < x < 4.71238 \dots + 2 πn

Solution steps

sin (2 x) > 6 cos (x)

Move

6 cos (x)

to the left side

sin (2 x) > 6 cos (x)

Subtract

6 cos (x)

from both sides

sin (2 x) - 6 cos (x) > 6 cos (x) - 6 cos (x)

sin (2 x) - 6 cos (x) > 0

sin (2 x) - 6 cos (x) > 0

Use the following identity:

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

2 cos (x) sin (x) - 6 cos (x) > 0

Periodicity of

2 cos (x) sin (x) - 6 cos (x) : 2 π

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

2 cos (x) sin (x), 6 cos (x)

Periodicity of

2 cos (x) sin (x) : π

2 cos (x) sin (x)

is composed of the following functions and periods:

cos (x)

with periodicity of

2 π

The compound periodicity is:

π

Periodicity of

6 cos (x) : 2 π

Periodicity of

a \cdot cos (b x + c) + d = \frac{periodicity of cos ( x )}{∣ b ∣}

Periodicity of

cos (x)

2 π

= \frac{2 π}{∣ 1 ∣}

Simplify

= 2 π

Combine periods:

π, 2 π

= 2 π

Factor

2 cos (x) sin (x) - 6 cos (x) : 2 cos (x) (sin (x) - 3)

2 cos (x) sin (x) - 6 cos (x)

Rewrite

- 6

3 \cdot 2

= 2 sin (x) cos (x) + 3 \cdot 2 cos (x)

Factor out common term

2 cos (x)

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

2 cos (x) (sin (x) - 3) > 0

To find the zeroes, set the inequality to zero

2 cos (x) (sin (x) - 3) = 0

Solve

2 cos (x) (sin (x) - 3) = 0

for

0 \leq x < 2 π

2 cos (x) (sin (x) - 3) = 0

Solving each part separately

cos (x) = 0 : x = \frac{π}{2} or x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x < 2 π

General solutions for

cos (x) = 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}

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

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

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

Combine all the solutions

\frac{π}{2} or \frac{3 π}{2}

The intervals between the zeros

0 < x < \frac{π}{2}, \frac{π}{2} < x < \frac{3 π}{2}, \frac{3 π}{2} < x < 2 π

Summarize in a table:

cos (x) sin (x) - 3 2 cos (x) (sin (x) - 3) x = 0 + - - 0 < x < \frac{π}{2} + - - x = \frac{π}{2} 0 - 0 \frac{π}{2} < x < \frac{3 π}{2} - - + x = \frac{3 π}{2} 0 - 0 \frac{3 π}{2} < x < 2 π + - - x = 2 π + - -

Identify the intervals that satisfy the required condition:

> 0

\frac{π}{2} < x < \frac{3 π}{2}

Apply the periodicity of

2 cos (x) sin (x) - 6 cos (x)

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

Popular Examples

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- cos (x) \leq - sin (2 x)

cos(y)>-1

cos (y) > - 1

2cos(x)+cos^2(x)>0

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tan(x)>-2/3

tan (x) > - \frac{2}{3}

sin(x)-cos(x)>= 1

sin (x) - cos (x) \geq 1