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

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

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solvefor x,arccos(x)-arccos(y/2)=a

solve for

x, arccos (x) - arccos (\frac{y}{2}) = a

x = cos (a + arccos (\frac{y}{2}))

Solution steps

arccos (x) - arccos (\frac{y}{2}) = a

Move

arccos (\frac{y}{2})

to the right side

arccos (x) - arccos (\frac{y}{2}) = a

Add

arccos (\frac{y}{2})

to both sides

arccos (x) - arccos (\frac{y}{2}) + arccos (\frac{y}{2}) = a + arccos (\frac{y}{2})

Simplify

arccos (x) = a + arccos (\frac{y}{2})

arccos (x) = a + arccos (\frac{y}{2})

Apply trig inverse properties

arccos (x) = a + arccos (\frac{y}{2})

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

x = cos (a + arccos (\frac{y}{2}))

x = cos (a + arccos (\frac{y}{2}))

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

tan (b x) = 5

sin (t + \frac{π}{4}) = \frac{1}{2}, π \leq t \leq \frac{7 π}{2}

tan (x) = 3007

2 sin (θ) - 3 = 0

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