# Solving an equation of one parameter over a range of values of that parameter [closed]

How can I use NSolve the following situation (to rectify a cycloid arc mapping with hashes etc.)?

For a specific value on the right-hand side, I use

NSolve[8 Sin[t/4]^2 == 4, t]


I want to solve over a range of values on the right-hand side, but

NSolve[8 Sin[t/4]^2 == n , t, {n, 1, 8, 1}]


doesn't work.

EDIT1:

nn=10;
Table[FindRoot[ X[s] == Xsmax  frac, {s , .5}], {frac, 0, 1,
1/nn}] // TableForm


The suggestion also works for repeated FindRoot as above.

## closed as off-topic by m_goldberg, MarcoB, dr.blochwave, ilian, bbgodfreyNov 25 '15 at 6:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – m_goldberg, MarcoB, dr.blochwave, ilian, bbgodfrey
If this question can be reworded to fit the rules in the help center, please edit the question.

• NSolve[{8 Sin[t/4]^2 == 4, 1 <= t <= 8}, t] – Jason B. Nov 23 '15 at 7:30
• I like to see 8 solutions, first line entry example.. – Narasimham Nov 23 '15 at 7:42
• I'm not sure what you need, Table[NSolve[8 Sin[t/4]^2 == n, t], {n, 8}]? – Kuba Nov 24 '15 at 8:52

## 2 Answers

My understanding from your question is that you want a solution set for each value of n going from 1 to 8 in steps of 1. If this is the case just use Table:

Table[NSolve[8 Sin[t/4]^2 == n, t], {n, 1, 8, 1}]


You may solve it exactly:

Solve[8 Sin[t/4]^2 == #, t] /. C -> 0 & /@ Range@8