What is the general solution for 6arccos(5x)=5pi ?

The general solution for 6arccos(5x)=5pi is x=-(sqrt(3))/(10)

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6arccos(5x)=5pi

6 arccos (5 x) = 5 π

x = - \frac{3}{10}

Solution steps

6 arccos (5 x) = 5 π

Divide both sides by

6

6 arccos (5 x) = 5 π

Divide both sides by

6

\frac{6 arccos ( 5 x )}{6} = \frac{5 π}{6}

Simplify

arccos (5 x) = \frac{5 π}{6}

arccos (5 x) = \frac{5 π}{6}

Apply trig inverse properties

arccos (5 x) = \frac{5 π}{6}

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

5 x = cos (\frac{5 π}{6})

cos (\frac{5 π}{6}) = - \frac{3}{2}

cos (\frac{5 π}{6})

Use the following trivial identity:

cos (\frac{5 π}{6}) = - \frac{3}{2}

cos (\frac{5 π}{6})

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{3}{2}

= - \frac{3}{2}

5 x = - \frac{3}{2}

5 x = - \frac{3}{2}

Solve

5 x = - \frac{3}{2} : x = - \frac{3}{10}

5 x = - \frac{3}{2}

Divide both sides by

5

5 x = - \frac{3}{2}

Divide both sides by

5

\frac{5 x}{5} = \frac{- \frac{3}{2}}{5}

Simplify

\frac{5 x}{5} = \frac{- \frac{3}{2}}{5}

Simplify

\frac{5 x}{5} : x

\frac{5 x}{5}

Divide the numbers:

\frac{5}{5} = 1

= x

Simplify

\frac{- \frac{3}{2}}{5} : - \frac{3}{10}

\frac{- \frac{3}{2}}{5}

Apply the fraction rule:

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

= - \frac{\frac{3}{2}}{5}

Apply the fraction rule:

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

\frac{\frac{3}{2}}{5} = \frac{3}{2 \cdot 5}

= - \frac{3}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= - \frac{3}{10}

x = - \frac{3}{10}

x = - \frac{3}{10}

x = - \frac{3}{10}

x = - \frac{3}{10}

sin (x) + sin (x) cot (x) = 0

arctan (θ) = 1

cos (2 θ) = - \frac{1}{2}, 0 \leq θ \leq 2 π

sin (2 x) = \frac{2}{5}

sin (2 x) + cos (x) = 0, 0 \leq x \leq 2 π

What is the general solution for 6arccos(5x)=5pi ?
The general solution for 6arccos(5x)=5pi is x=-(sqrt(3))/(10)