What is the general solution for solvefor x,z=arccos(x+y) ?

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

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solvefor x,z=arccos(x+y)

solve for

x, z = arccos (x + y)

x = cos (z) - y

Solution steps

z = arccos (x + y)

Switch sides

arccos (x + y) = z

Apply trig inverse properties

arccos (x + y) = z

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

x + y = cos (z)

x + y = cos (z)

Solve

x + y = cos (z) : x = cos (z) - y

x + y = cos (z)

Move

y

to the right side

x + y = cos (z)

Subtract

y

from both sides

x + y - y = cos (z) - y

Simplify

x = cos (z) - y

x = cos (z) - y

x = cos (z) - y

3 tan (x) - 2 cot (x) = 5

tan (θ) = \frac{\frac{3}{1} - \frac{2}{5}}{1 + \frac{3}{1} \cdot \frac{2}{5}}

tan (t) = 14

cos (A) = \frac{5}{8}

sin (9 x + 40) = cos (- 5 x + 42)

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