What is the general solution for solvefor y,arccos(y/2)=5log_{10}(x/5) ?

The general solution for solvefor y,arccos(y/2)=5log_{10}(x/5) is y=2cos(5log_{10}(x/5))

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solvefor y,arccos(y/2)=5log_{10}(x/5)

solve for

y, arccos (\frac{y}{2}) = 5 lo g_{10} (\frac{x}{5})

y = 2 cos (5 lo g_{10} (\frac{x}{5}))

Solution steps

arccos (\frac{y}{2}) = 5 lo g_{10} (\frac{x}{5})

Apply trig inverse properties

arccos (\frac{y}{2}) = 5 lo g_{10} (\frac{x}{5})

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

\frac{y}{2} = cos (5 lo g_{10} (\frac{x}{5}))

\frac{y}{2} = cos (5 lo g_{10} (\frac{x}{5}))

Solve

\frac{y}{2} = cos (5 lo g_{10} (\frac{x}{5})) : y = 2 cos (5 lo g_{10} (\frac{x}{5}))

\frac{y}{2} = cos (5 lo g_{10} (\frac{x}{5}))

Multiply both sides by

2

\frac{y}{2} = cos (5 lo g_{10} (\frac{x}{5}))

Multiply both sides by

2

\frac{2 y}{2} = 2 cos (5 lo g_{10} (\frac{x}{5}))

Simplify

y = 2 cos (5 lo g_{10} (\frac{x}{5}))

y = 2 cos (5 lo g_{10} (\frac{x}{5}))

y = 2 cos (5 lo g_{10} (\frac{x}{5}))

sin^{22} (x) = \frac{1}{4}

3 sin (2 x) - cos (x) = 0

2 cos^{2} (x) + tan^{2} (x) = 2

sin^{3} (o) = 4 sin (o) sin^{2} (o) sin^{4} (o)

cos^{2} (x) = - 4

What is the general solution for solvefor y,arccos(y/2)=5log_{10}(x/5) ?
The general solution for solvefor y,arccos(y/2)=5log_{10}(x/5) is y=2cos(5log_{10}(x/5))