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Popular >

sin(θ)sec(θ)>0\land sin(θ)<4

Solution

sin (θ) sec (θ) > 0 and sin (θ) < 4

Solution

πn < θ < \frac{π}{2} + πn

Interval Notation

(πn, \frac{π}{2} + πn)

Decimal Notation

πn < θ < 1.57079 \dots + πn

Solution steps

sin (θ) sec (θ) > 0 and sin (θ) < 4

sin (θ) sec (θ) > 0 : πn < θ < \frac{π}{2} + πn

sin (θ) sec (θ) > 0

Periodicity of

sin (θ) sec (θ) : π

sin (θ) sec (θ)

is composed of the following functions and periods:

sin (θ)

with periodicity of

2 π

The compound periodicity is:

= π

Express with sin, cos

sin (θ) sec (θ) > 0

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

sin (θ) \frac{1}{cos ( θ )} > 0

sin (θ) \frac{1}{cos ( θ )} > 0

Simplify

sin (θ) \frac{1}{cos ( θ )} : \frac{sin ( θ )}{cos ( θ )}

sin (θ) \frac{1}{cos ( θ )}

Multiply fractions:

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

= \frac{1 \cdot sin ( θ )}{cos ( θ )}

Multiply:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ )}{cos ( θ )}

\frac{sin ( θ )}{cos ( θ )} > 0

Find the zeroes and undifined points of

\frac{sin ( θ )}{cos ( θ )}

for

0 \leq θ < π

To find the zeroes, set the inequality to zero

\frac{sin ( θ )}{cos ( θ )} = 0

\frac{sin ( θ )}{cos ( θ )} = 0, 0 \leq θ < π : θ = 0

\frac{sin ( θ )}{cos ( θ )} = 0, 0 \leq θ < π

Rewrite using trig identities

\frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

= tan (θ)

tan (θ) = 0

General solutions for

tan (θ) = 0

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}

θ = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

Solutions for the range

0 \leq θ < π

θ = 0

Find the undefined points:

θ = \frac{π}{2}

Find the zeros of the denominator

cos (θ) = 0

General solutions for

cos (θ) = 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}

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

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

Solutions for the range

0 \leq θ < π

θ = \frac{π}{2}

0, \frac{π}{2}

Identify the intervals

0 < θ < \frac{π}{2}, \frac{π}{2} < θ < π

Summarize in a table:

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

Identify the intervals that satisfy the required condition:

> 0

0 < θ < \frac{π}{2}

Apply the periodicity of

sin (θ) sec (θ)

πn < θ < \frac{π}{2} + πn

sin (θ) < 4 :

True for all

θ \in R

sin (θ) < 4

Range of

sin (θ) : - 1 \leq sin (θ) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (θ) \leq 1

- 1 \leq sin (θ) \leq 1

sin (θ) < 4 and - 1 \leq sin (θ) \leq 1 : - 1 \leq sin (θ) \leq 1

Let

y = sin (θ)

Combine the intervals

y < 4 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y < 4 and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y < 4

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all θ

True for all θ \in R

Combine the intervals

πn < θ < \frac{π}{2} + πn and True for all θ \in R

Merge Overlapping Intervals

πn < θ < \frac{π}{2} + πn

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csc (x) = \frac{- 13}{2} and tan (x) > 0

-1<sin(x)<1

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-2<= 2cos(3x+5)<= 2

- 2 \leq 2 cos (3 x + 5) \leq 2

arccos(-0.83)180<θ<270

arccos (- 0.83) 180 < θ < 270

-1/2 <sin(x)<1

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