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.
sin(bx+c) = 1/2
bx+c = pi/6 or 5pi/6
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.
To find the number of solutions for the equation asin(bx+c)+d=d+a/2 in the interval of 0 < x < 360, you can follow these steps:
Step 1: Simplify the equation
Start by canceling out the "d" terms. The equation becomes:
asin(bx+c) = a/2
Step 2: Isolate the sine term
Divide both sides of the equation by "a":
sin(bx+c) = 1/2a
Step 3: Determine the possible values for the sine term
Sine function values range from -1 to 1. Therefore, the possible values for the sine term (1/2a) are from -1/(2a) to 1/(2a).
Step 4: Determine the range for the angle (bx+c)
Since the range of the sine function is from -1 to 1, the angle (bx+c) should fall within the range of -π/2 to π/2.
Step 5: Solve for "x"
To find the number of solutions in the interval 0 < x < 360, divide the range of the angle by the positive integer "b". This will give us the number of positive solutions within the interval.
Number of solutions = (π/2 - (-π/2)) / b
= π / b
So, the number of solutions in the interval 0 < x < 360 should be approximately π / b, where "b" is a positive integer and a doesn't equal zero.