What is the general solution for arccos(2x-1)= pi/6 ?

The general solution for arccos(2x-1)= pi/6 is x=(sqrt(3))/4+1/2

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arccos(2x-1)= pi/6

arccos (2 x - 1) = \frac{π}{6}

x = \frac{3}{4} + \frac{1}{2}

Solution steps

arccos (2 x - 1) = \frac{π}{6}

Apply trig inverse properties

arccos (2 x - 1) = \frac{π}{6}

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

2 x - 1 = 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}

2 x - 1 = \frac{3}{2}

2 x - 1 = \frac{3}{2}

Solve

2 x - 1 = \frac{3}{2} : x = \frac{3}{4} + \frac{1}{2}

2 x - 1 = \frac{3}{2}

Move

1

to the right side

2 x - 1 = \frac{3}{2}

Add

1

to both sides

2 x - 1 + 1 = \frac{3}{2} + 1

Simplify

2 x = \frac{3}{2} + 1

2 x = \frac{3}{2} + 1

Divide both sides by

2

2 x = \frac{3}{2} + 1

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{3}{2}}{2} + \frac{1}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{3}{2}}{2} + \frac{1}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{3}{2}}{2} + \frac{1}{2} : \frac{3}{4} + \frac{1}{2}

\frac{\frac{3}{2}}{2} + \frac{1}{2}

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

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

Apply the fraction rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3}{4}

= \frac{3}{4} + \frac{1}{2}

x = \frac{3}{4} + \frac{1}{2}

x = \frac{3}{4} + \frac{1}{2}

x = \frac{3}{4} + \frac{1}{2}

x = \frac{3}{4} + \frac{1}{2}

sin (θ) - cos (\frac{θ}{2}) = 0

tan (x) = \frac{195}{867}

x, cos (q x) = sin (r x)

sin (21 6^{\circ}) = sin (θ)

cos (θ) = - 0.2

What is the general solution for arccos(2x-1)= pi/6 ?
The general solution for arccos(2x-1)= pi/6 is x=(sqrt(3))/4+1/2