posted by Anonymous
2sin^2(x )+38 cos(x) + 2.6=0
Change Sin^2(x) into 1-cos^2(x), substitute it in, tidy it up and you'll get a quadratic of the form
aCos^2(x) + bCos(x) + c = 0
Then let, say, y = Cos(x) then and you get
ay^2 + by + c = 0
which is a more normal quadratic. Solve that for y and put your answer equal to Cos(x). There will be 2 possible values of y probably and 2 possible values for Cos (x) for each value of y. In total, 4 answers.