PLEASE HELP ME If 0 < z ≤ 90 and sin(9z − 1) = cos(6z + 1), what is the value of z? z = 3 z = 4 z = 5 z = 6

  z = 6

Step-by-step explanation:

We know that ...

  sin(x) = cos(90 -x)

Substituting (9z-1) for x, this is ...

  sin(9z -1) = cos(90 -(9z -1))

But we also are given ...

  sin(9z -1) = cos(6z +1)

Equating the arguments of the cosine function, we have ...

  90 -(9z -1) = 6z +1

  90 = 15z . . . . . . . . . add (9z-1) to both sides

  6 = z . . . . . . . . . . . . divide by 15

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Comment on the graph

The attached graph shows 5 solutions in the domain of interest. These come from the fact that the relation we used is actually ...

  sin(x) = cos(90 +360k -x)  . . . . .  for any integer k

Then the above equation becomes ...

  90 +360k = 15z

  6 +24k = z . . . . . . . . . for any integer k

The sine and cosine functions also enjoy the relation ...

  sin(x) = cos(x -90)

  sin(9z -1) = cos(9z -1 -90) = cos(6z +1)

  3z = 92 . . . . . equating arguments of cos( ) and adding 91-6z

  z = 30 2/3

answer: -2+√x,-2-√x

i hope this , i don't know if you're looking for a pair or not.