Question
solved "problems" of 2sin^2-5sinx-3= 2π?
Answers
oobleck
what an unusual problem. As written, you (sort of) have
2sin^2x - 5sinx - 3 = 2π
so, per the quadratic formula, that yields
sinx = (5±√(25+8(3+2π))/4
But those are both too big, since |sinx| <= 1
I suspect you have mangled the problem.
2sin^2x - 5sinx - 3 = 2π
so, per the quadratic formula, that yields
sinx = (5±√(25+8(3+2π))/4
But those are both too big, since |sinx| <= 1
I suspect you have mangled the problem.