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Popular >

2cos^2(x)+sin(2x)<= 0

Solution

2 cos^{2} (x) + sin (2 x) \leq 0

Solution

\frac{π}{2} + πn \leq x \leq \frac{3 π}{4} + πn

Interval Notation

[\frac{π}{2} + πn, \frac{3 π}{4} + πn]

Decimal Notation

1.57079 \dots + πn \leq x \leq 2.35619 \dots + πn

Solution steps

2 cos^{2} (x) + sin (2 x) \leq 0

Use the following identity:

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

2 cos (x) sin (x) + 2 cos^{2} (x) \leq 0

Periodicity of

2 cos (x) sin (x) + 2 cos^{2} (x) : π

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

2 cos (x) sin (x), 2 cos^{2} (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

2 cos^{2} (x) : π

Periodicity of

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

if n is even

Periodicity of

cos (x) : 2 π

Periodicity of

cos (x)

2 π

= 2 π

\frac{2 π}{2}

Simplify

π

Combine periods:

π, π

= π

Factor

2 cos (x) sin (x) + 2 cos^{2} (x) : 2 cos (x) (sin (x) + cos (x))

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

Apply exponent rule:

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

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

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

Factor out common term

2 cos (x)

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

2 cos (x) (sin (x) + cos (x)) \leq 0

To find the zeroes, set the inequality to zero

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

Solve

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

for

0 \leq x < π

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

Solving each part separately

cos (x) = 0 : x = \frac{π}{2}

cos (x) = 0, 0 \leq x < 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 < π

x = \frac{π}{2}

sin (x) + cos (x) = 0 : x = \frac{3 π}{4}

sin (x) + cos (x) = 0, 0 \leq x < π

Rewrite using trig identities

sin (x) + cos (x) = 0

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

tan (x) + 1 = 0

tan (x) + 1 = 0

Move

1

to the right side

tan (x) + 1 = 0

Subtract

1

from both sides

tan (x) + 1 - 1 = 0 - 1

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (x) = - 1

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}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Solutions for the range

0 \leq x < π

x = \frac{3 π}{4}

Combine all the solutions

\frac{π}{2} or \frac{3 π}{4}

The intervals between the zeros

0 < x < \frac{π}{2}, \frac{π}{2} < x < \frac{3 π}{4}, \frac{3 π}{4} < x < π

Summarize in a table:

cos (x) sin (x) + cos (x) 2 cos (x) (sin (x) + cos (x)) x = 0 + + + 0 < x < \frac{π}{2} + + + x = \frac{π}{2} 0 + 0 \frac{π}{2} < x < \frac{3 π}{4} - + - x = \frac{3 π}{4} - 00 \frac{3 π}{4} < x < π - - + x = π - - +

Identify the intervals that satisfy the required condition:

\leq 0

x = \frac{π}{2} or \frac{π}{2} < x < \frac{3 π}{4} or x = \frac{3 π}{4}

Merge Overlapping Intervals

\frac{π}{2} \leq x < \frac{3 π}{4} or x = \frac{3 π}{4}

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

x = \frac{π}{2}

\frac{π}{2} < x < \frac{3 π}{4}

\frac{π}{2} \leq x < \frac{3 π}{4}

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

\frac{π}{2} \leq x < \frac{3 π}{4}

x = \frac{3 π}{4}

\frac{π}{2} \leq x \leq \frac{3 π}{4}

\frac{π}{2} \leq x \leq \frac{3 π}{4}

Apply the periodicity of

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

\frac{π}{2} + πn \leq x \leq \frac{3 π}{4} + πn

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