sin(x)*sin2(x) = 1 so sin3(x) = 1 so that sin(x) = cuberoot(1) = 1 then x = pi/2 + n*pi where n is an integer.
You cannot. sin(n)/n is an expression, not an equation. An expression cannot be solved.
It is impossible to solve for n if there's nothing to equate it with. No could be any number each except for -4. All we know is n = n
you are saying 1.5 is equal to n over and 1.5 is equal to 1 over 6 first of all 1.5 is not euqal to one over six. but...if 1.5 = (n/3) then n=4.5
If sin2(theta) = 0, then theta is N pi, N being any integer
I was told it is x(n) = 10*sin(2*pi*fo*Ts(n-1)) + 8*sin(2*pi*2*fo*Ts(n-1)) + 6*sin(2*pi*3*fo*Ts(n-1))
(1-(SIN(n)^2))^0.5
Unfortunately, the browser used for posting questions is hopelessly inadequate for mathematics: it strips away most symbols. All that we can see is "sin(-1)sin((5pi )(7))". From that it is not at all clear what the missing symbols (operators) between (5pi ) and (7) might be. There is, therefore no sensible answer. It makes little sense for me to try and guess - I may as well make up my own questions and answer them!All that I can tell you that the principal sin-1 is the inverse for sin over the domain (-pi/2, pi/2). Thus sin-1(sin(x) = x where -pi/2 < x
n=-4/3
2
n+1=n solve for n.
6(n - 4)(n + 1) n = 4, -1