What is the general solution for arccos(2x)=0 ?

The general solution for arccos(2x)=0 is x= 1/2

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arccos(2x)=0

arccos (2 x) = 0

x = \frac{1}{2}

Solution steps

arccos (2 x) = 0

Apply trig inverse properties

arccos (2 x) = 0

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

2 x = cos (0)

cos (0) = 1

cos (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

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}

= 1

= 1

2 x = 1

2 x = 1

Solve

2 x = 1 : x = \frac{1}{2}

2 x = 1

Divide both sides by

2

2 x = 1

Divide both sides by

2

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

Simplify

x = \frac{1}{2}

x = \frac{1}{2}

x = \frac{1}{2}

cos (x + 3 0^{\circ}) = 0

cos (2 x) - sin^{2} (x) - \frac{1}{4} = 0

tan (3 x + 1) = 1

- cos (2 x) = 2 cos (2 x) - 1

sin (x) = 23

What is the general solution for arccos(2x)=0 ?
The general solution for arccos(2x)=0 is x= 1/2