sin(bx+c) = 1/2
bx+c = pi/6 or 5pi/6
How many solutiond in the interval of 0 < x < 360 should you expect for the equation asin(bx+c)+d=d+a/2
A does not equal to 0 and b is a positive integer. I have cancelled the d’s and i understand the range will be from a to a-, but i dont get how to get the solution.
2 answers
consider sin(x) = v for some v. It has two solutions in [0,360) because sin(x) is positive in QI and QII, and negative in QIII and QIV.
since sin(bx) has period 360/b, expect two solutions in the interval [0,360/b)
Or, 2b solutions in [0,360)
Not sure why you specify 0 < x < 360, but if so, then any solution where x=0 must be discarded.
since sin(bx) has period 360/b, expect two solutions in the interval [0,360/b)
Or, 2b solutions in [0,360)
Not sure why you specify 0 < x < 360, but if so, then any solution where x=0 must be discarded.