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# C3 Trig help watch

1. Find the values of R and a, where R > 0 and 0 < a < 90 for which 5/(8cosx - 6sinx) =Rsec(x+a)

so far I know that it can be changed to R/Cos(x+a) but I'm not too sure where to go from there
2. (Original post by not_lucas1)
Find the values of R and a, where R > 0 and 0 < a < 90 for which 5/(8cosx - 6sinx) =Rsec(x+a)

so far I know that it can be changed to R/Cos(x+a) but I'm not too sure where to go from there
If you take the inverse of both sides, you want to find which is a standard problem.

In particular: , can you find ?
3. (Original post by Zacken)
If you take the inverse of both sides, you want to find which is a standard problem.

In particular: , can you find ?
I ended up with 24/25 Cos(x+53.13) which i changed to 25/24 Sec(x+53.13)
not sure if its right
4. (Original post by not_lucas1)
I ended up with 24/25 Cos(x+53.13) which i changed to 25/24 Sec(x+53.13)
not sure if its right
Don't think it is.
5. well 1/cosx = secx so if youve got 5/Zcos(x +a) then your answer will be 5/Z sec (x + a), where 5/Z is equal to R.

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