What is the general solution for solvefor x,y=5arccos(2x) ?

The general solution for solvefor x,y=5arccos(2x) is x=(cos(y/5))/2

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solvefor x,y=5arccos(2x)

solve for

x, y = 5 arccos (2 x)

x = \frac{cos ( \frac{y}{5} )}{2}

Solution steps

y = 5 arccos (2 x)

Switch sides

5 arccos (2 x) = y

Divide both sides by

5

5 arccos (2 x) = y

Divide both sides by

5

\frac{5 arccos ( 2 x )}{5} = \frac{y}{5}

Simplify

arccos (2 x) = \frac{y}{5}

arccos (2 x) = \frac{y}{5}

Apply trig inverse properties

arccos (2 x) = \frac{y}{5}

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

2 x = cos (\frac{y}{5})

2 x = cos (\frac{y}{5})

Solve

2 x = cos (\frac{y}{5}) : x = \frac{cos ( \frac{y}{5} )}{2}

2 x = cos (\frac{y}{5})

Divide both sides by

2

2 x = cos (\frac{y}{5})

Divide both sides by

2

\frac{2 x}{2} = \frac{cos ( \frac{y}{5} )}{2}

Simplify

x = \frac{cos ( \frac{y}{5} )}{2}

x = \frac{cos ( \frac{y}{5} )}{2}

x = \frac{cos ( \frac{y}{5} )}{2}

sin (3 x) = \frac{2}{2}, 0 \leq x \leq 2 π

\frac{cot ( x )}{cos ( x )} + \frac{1}{sin ( x )} = 1

3 sin (θ) = sin (θ) - 2

sec^{2} (x) - 2 sec (x) + 1 = 0

cos (θ) = sin (θ - \frac{π}{3})

What is the general solution for solvefor x,y=5arccos(2x) ?
The general solution for solvefor x,y=5arccos(2x) is x=(cos(y/5))/2