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Popular Trigonometry >

(2sin(2x)+sqrt(2))*tan(x)<0

Solution

(2 sin (2 x) + 2) \cdot tan (x) < 0

Solution

\frac{π}{2} + πn < x < \frac{5 π}{8} + πn or \frac{7 π}{8} + πn < x < π + πn

Interval Notation

(\frac{π}{2} + πn, \frac{5 π}{8} + πn) \cup (\frac{7 π}{8} + πn, π + πn)

Decimal

1.57079 \dots + πn < x < 1.96349 \dots + πn or 2.74889 \dots + πn < x < 3.14159 \dots + πn

Solution steps

(2 sin (2 x) + 2) tan (x) < 0

Periodicity of

(2 sin (2 x) + 2) tan (x) : π

(2 sin (2 x) + 2) tan (x)

is composed of the following functions and periods:

sin (2 x)

with periodicity of

\frac{2 π}{2}

The compound periodicity is:

= π

Express with sin, cos

(2 sin (2 x) + 2) tan (x) < 0

Use the basic trigonometric identity:

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

(2 sin (2 x) + 2) \frac{sin ( x )}{cos ( x )} < 0

(2 sin (2 x) + 2) \frac{sin ( x )}{cos ( x )} < 0

Simplify

(2 sin (2 x) + 2) \frac{sin ( x )}{cos ( x )} : \frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )}

(2 sin (2 x) + 2) \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

= \frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )}

\frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )} < 0

Find the zeroes and undifined points of

\frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )}

for

0 \leq x < π

To find the zeroes, set the inequality to zero

\frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )} = 0

\frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )} = 0, 0 \leq x < π : x = 0, x = \frac{5 π}{8}, x = \frac{7 π}{8}

\frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )} = 0, 0 \leq x < π

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (x) (2 sin (2 x) + 2) = 0

Solving each part separately

sin (x) = 0 or 2 sin (2 x) + 2 = 0

sin (x) = 0, 0 \leq x < π : x = 0

sin (x) = 0, 0 \leq x < π

General solutions for

sin (x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Solutions for the range

0 \leq x < π

x = 0

2 sin (2 x) + 2 = 0, 0 \leq x < π : x = \frac{5 π}{8}, x = \frac{7 π}{8}

2 sin (2 x) + 2 = 0, 0 \leq x < π

Move

2

to the right side

2 sin (2 x) + 2 = 0

Subtract

2

from both sides

2 sin (2 x) + 2 - 2 = 0 - 2

Simplify

2 sin (2 x) = - 2

2 sin (2 x) = - 2

Divide both sides by

2

2 sin (2 x) = - 2

Divide both sides by

2

\frac{2 sin ( 2 x )}{2} = \frac{- 2}{2}

Simplify

sin (2 x) = - \frac{2}{2}

sin (2 x) = - \frac{2}{2}

General solutions for

sin (2 x) = - \frac{2}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = \frac{5 π}{4} + 2 πn, 2 x = \frac{7 π}{4} + 2 πn

2 x = \frac{5 π}{4} + 2 πn, 2 x = \frac{7 π}{4} + 2 πn

Solve

2 x = \frac{5 π}{4} + 2 πn : x = \frac{5 π}{8} + πn

2 x = \frac{5 π}{4} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{4} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2} : \frac{5 π}{8} + πn

\frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{4}}{2} = \frac{5 π}{8}

\frac{\frac{5 π}{4}}{2}

Apply the fraction rule:

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

= \frac{5 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{5 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

Solve

2 x = \frac{7 π}{4} + 2 πn : x = \frac{7 π}{8} + πn

2 x = \frac{7 π}{4} + 2 πn

Divide both sides by

2

2 x = \frac{7 π}{4} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2} : \frac{7 π}{8} + πn

\frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{7 π}{4}}{2} = \frac{7 π}{8}

\frac{\frac{7 π}{4}}{2}

Apply the fraction rule:

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

= \frac{7 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{7 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{7 π}{8} + πn

x = \frac{7 π}{8} + πn

x = \frac{7 π}{8} + πn

x = \frac{7 π}{8} + πn

x = \frac{5 π}{8} + πn, x = \frac{7 π}{8} + πn

Solutions for the range

0 \leq x < π

x = \frac{5 π}{8}, x = \frac{7 π}{8}

Combine all the solutions

x = 0, x = \frac{5 π}{8}, x = \frac{7 π}{8}

Find the undefined points:

x = \frac{π}{2}

Find the zeros of the denominator

cos (x) = 0

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 < π

x = \frac{π}{2}

0, \frac{π}{2}, \frac{5 π}{8}, \frac{7 π}{8}

Identify the intervals

0 < x < \frac{π}{2}, \frac{π}{2} < x < \frac{5 π}{8}, \frac{5 π}{8} < x < \frac{7 π}{8}, \frac{7 π}{8} < x < π

Summarize in a table:

sin (x) 2 sin (2 x) + 2 cos (x) \frac{s i n ( x ) ( 2 s i n ( 2 x ) + 2 )}{c o s ( x )} x = 0 0 + + 0 0 < x < \frac{π}{2} + + + + x = \frac{π}{2} + + 0 Undefined \frac{π}{2} < x < \frac{5 π}{8} + + - - x = \frac{5 π}{8} + 0 - 0 \frac{5 π}{8} < x < \frac{7 π}{8} + - - + x = \frac{7 π}{8} + 0 - 0 \frac{7 π}{8} < x < π + + - - x = π 0 + - 0

Identify the intervals that satisfy the required condition:

< 0

\frac{π}{2} < x < \frac{5 π}{8} or \frac{7 π}{8} < x < π

Apply the periodicity of

(2 sin (2 x) + 2) tan (x)

\frac{π}{2} + πn < x < \frac{5 π}{8} + πn or \frac{7 π}{8} + πn < x < π + πn

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