What is the general solution for 6arccos(4x)=pi ?

The general solution for 6arccos(4x)=pi is x=(sqrt(3))/8

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6arccos(4x)=pi

6 arccos (4 x) = π

x = \frac{3}{8}

Solution steps

6 arccos (4 x) = π

Divide both sides by

6

6 arccos (4 x) = π

Divide both sides by

6

\frac{6 arccos ( 4 x )}{6} = \frac{π}{6}

Simplify

arccos (4 x) = \frac{π}{6}

arccos (4 x) = \frac{π}{6}

Apply trig inverse properties

arccos (4 x) = \frac{π}{6}

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

4 x = cos (\frac{π}{6})

cos (\frac{π}{6}) = \frac{3}{2}

cos (\frac{π}{6})

Use the following trivial identity:

cos (\frac{π}{6}) = \frac{3}{2}

cos (\frac{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}

4 x = \frac{3}{2}

4 x = \frac{3}{2}

Solve

4 x = \frac{3}{2} : x = \frac{3}{8}

4 x = \frac{3}{2}

Divide both sides by

4

4 x = \frac{3}{2}

Divide both sides by

4

\frac{4 x}{4} = \frac{\frac{3}{2}}{4}

Simplify

\frac{4 x}{4} = \frac{\frac{3}{2}}{4}

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{3}{2}}{4} : \frac{3}{8}

\frac{\frac{3}{2}}{4}

Apply the fraction rule:

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

= \frac{3}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{3}{8}

x = \frac{3}{8}

x = \frac{3}{8}

x = \frac{3}{8}

x = \frac{3}{8}

- 4 sin (x) cos (x) = 2 sin (x)

7 sin^{2} (θ) + 16 sin (θ) + 9 = 0

3 tan (\frac{3 π}{4} (x + \frac{3 π}{2})) = 1

tan (x) = \frac{10}{4}

3 tan^{2} (x) + 11 tan (x) + 8 = 0

What is the general solution for 6arccos(4x)=pi ?
The general solution for 6arccos(4x)=pi is x=(sqrt(3))/8