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1/3 cosh(3x)0<= x<= ln(6^{(1/3)})

Solution

\frac{1}{3} cosh (3 x) 0 \leq x \leq ln (6^{(\frac{1}{3})})

0 \leq x \leq \frac{1}{3} ln (6)

Interval Notation

[0, \frac{1}{3} ln (6)]

Decimal Notation

0 \leq x \leq 0.59725 \dots

Solution steps

\frac{1}{3} cosh (3 x) \cdot 0 \leq x \leq ln (6^{(\frac{1}{3})})

a \leq u \leq b

then

a \leq u and u \leq b

\frac{1}{3} cosh (3 x) \cdot 0 \leq x and x \leq ln (6^{(\frac{1}{3})})

\frac{1}{3} cosh (3 x) \cdot 0 \leq x : x \geq 0

\frac{1}{3} cosh (3 x) \cdot 0 \leq x

Apply rule

0 \cdot a = 0

0 \leq x

Move

0

to the right side

0 \leq x

Simplify

\leq x

\leq x

Move

x

to the left side

\leq x

Subtract

x

from both sides

- x \leq x - x

Simplify

- x \leq 0

- x \leq 0

Multiply both sides by

- 1

- x \leq 0

Multiply both sides by -1 (reverse the inequality)

(- x) (- 1) \geq 0 \cdot (- 1)

Simplify

x \geq 0

x \geq 0

x \leq ln (6^{\frac{1}{3}}) : x \leq \frac{1}{3} ln (6)

x \leq ln (6^{\frac{1}{3}})

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

x \leq \frac{1}{3} ln (6)

Combine the intervals

x \geq 0 and x \leq \frac{1}{3} ln (6)

Merge Overlapping Intervals

x \geq 0 and x \leq \frac{1}{3} ln (6)

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

x \geq 0

and

x \leq \frac{1}{3} ln (6)

0 \leq x \leq \frac{1}{3} ln (6)

0 \leq x \leq \frac{1}{3} ln (6)

\frac{1}{2} \leq cos (x) \leq \frac{3}{2}

sin (θ) = - \frac{5}{9} and cos (θ) > 0

- 1 < sin (\frac{x}{6}) < 1

- 3 < tan (x) < 1

- 1 \leq cos (2 n) \leq 1