What is the general solution for solvefor x,arccos(e^{-x})= pi/3 ?

The general solution for solvefor x,arccos(e^{-x})= pi/3 is x=ln(2)

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solvefor x,arccos(e^{-x})= pi/3

solve for

x, arccos (e^{- x}) = \frac{π}{3}

x = ln (2)

Solution steps

arccos (e^{- x}) = \frac{π}{3}

Apply trig inverse properties

arccos (e^{- x}) = \frac{π}{3}

arccos (x) = a \Rightarrow x = cos (a)

e^{- x} = cos (\frac{π}{3})

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

Use the following trivial identity:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

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}

= \frac{1}{2}

= \frac{1}{2}

e^{- x} = \frac{1}{2}

e^{- x} = \frac{1}{2}

Solve

e^{- x} = \frac{1}{2} : x = ln (2)

e^{- x} = \frac{1}{2}

Apply exponent rules

e^{- x} = \frac{1}{2}

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

ln (e^{- x}) = ln (\frac{1}{2})

Apply log rule:

ln (e^{a}) = a

ln (e^{- x}) = - x

- x = ln (\frac{1}{2})

- x = ln (\frac{1}{2})

Solve

- x = ln (\frac{1}{2}) : x = ln (2)

- x = ln (\frac{1}{2})

Divide both sides by

- 1

- x = ln (\frac{1}{2})

Divide both sides by

- 1

\frac{- x}{- 1} = \frac{ln ( \frac{1}{2} )}{- 1}

Simplify

\frac{- x}{- 1} = \frac{ln ( \frac{1}{2} )}{- 1}

Simplify

\frac{- x}{- 1} : x

\frac{- x}{- 1}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{x}{1}

Apply rule

\frac{a}{1} = a

= x

Simplify

\frac{ln ( \frac{1}{2} )}{- 1} : ln (2)

\frac{ln ( \frac{1}{2} )}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{ln ( \frac{1}{2} )}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{ln ( \frac{1}{2} )}{1} = ln (\frac{1}{2})

= - ln (\frac{1}{2})

Apply log rule:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - (- ln (2))

Apply rule

- (- a) = a

= ln (2)

x = ln (2)

x = ln (2)

x = ln (2)

x = ln (2)

x = ln (2)

2 cos^{2} (4 x + 1) = 0

x, 2 sin (x) = - 1

t, f = a cos (\frac{t π}{6} - \frac{11 π}{12})

- 9 cos^{2} (θ) + 9 = 16 sin (θ) - 7, 0 \leq θ < 36 0^{\circ}

\frac{cos ( x )}{sin ^{2} ( x )} = \frac{2}{3}

What is the general solution for solvefor x,arccos(e^{-x})= pi/3 ?
The general solution for solvefor x,arccos(e^{-x})= pi/3 is x=ln(2)