Are you using the . in siny.cosy as a multiplication sign?
If so, then
siny = sinycosy
cosy = 1
and cos^2 y + cosy = p
1+1 = p
p = 2
Am I missing something here??
Looks too easy, did you have a typo?
It is given that cos^2 y + cos y = p . Determine the value of p if sin y = sin y . Cos y
2 answers
of course , if siny = 0, then we have cosy = 1 or -1.
You can try those for other solutions.
You can try those for other solutions.