# Determine the # of solutions to the equation

hamface
THE QUESTION:

determine the # of solutions to the equation...

sinax=b

for the domain 0 (less than or equal to) x (less than or equal to) 2pi, for the values of a. a is a pos. integer and b is a real #.

Then it gives examples:

value of a ..... value of b
......1 ..... ........... 0
...... 2 ..... ........... 0.2
......3 ..... ............ -0.2
..... 4 ..... .......... 1
.....0.5 ..... .......... 0.5
...... 1 ..... ............ 2

sin (ax) = b

From a = 1 and b = 0:
sin x = 0
x = {0, pi, 2(pi)}

From a = 2 and b = 0.2:
sin (2x) = 0.2
2x = approximately {0.2014, 2.9402, 6.4845, 9.2234}
x = approximately {0.1007, 1.4701, 3.2423, 4.6117}

From a = 3 and b = -0.2:
sin (3x) = -0.2
3x = approximately {3.3430, 6.0818, 9.6261, 12.3650, 15.9093, 18.6482}
x = approximately {1.1143, 2.0273, 3.2087, 4.1217, 5.3031, 6.2161}

From a = 4 and b = 1:
sin (4x) = 1
4x = {(1/2)(pi), (5/2)(pi), (9/2)(pi), (13/2)(pi)}
x = {(1/8)(pi), (5/8)(pi), (9/8)(pi), (13/8)(pi)}

...But I am not sure how to explain what I did out. And also, I didn't do the last two options, because they messed me up.

Can anyone help?

I need help.

Originally posted by hamface
THE for the values of a. a is a pos. integer and b is a real #.

Then it gives examples:

value of a ..... value of b
......1 ..... ........... 0
...... 2 ..... ........... 0.2
......3 ..... ............ -0.2
..... 4 ..... .......... 1
.....0.5 ..... .......... 0.5
...... 1 ..... ............ 2

0.5 isn't an integer.
Besides I don't understand the question. Are a and b constant? If not, then of course there are either 1 or 2 solutions for any pair of a and b. But from the examples above, it appears that a and b are not constant. In which case, what do you mean by number of solutions? Do you mean the number of pairs of a,b such that the equality holds for some x between 0 and 2pi? Do you mean the number of x values that saitify the equation for at least one a,b pair?

Also,
___
From a = 4 and b = 1:
sin (4x) = 1
4x = {(1/2)(pi), (5/2)(pi), (9/2)(pi), (13/2)(pi)}
x = {(1/8)(pi), (5/8)(pi), (9/8)(pi), (13/8)(pi)}
___

5/2(pi) is not in the domain. Nor is 13/2(pi). You said 0<x<2pi or x=0 or x=2pi

ks
You need to consider a lot of different cases:

If b < -1 or b > 1, there are no solutions regardless of the value of a.

For other possible values of b, first figure out the number of solutions for sin x = b (a = 1).

There are 3 possible cases to consider:

If b = -1 or 1, then there is only one solution to this equation.

If 0 < b < 1 or -1 < b < 0, there are 2 possible solutions to this equation.

If b = 0, there are 3 solutions.

For a > 1, if x is to be between 0 and 2Pi inclusive then ax must be between 0 and 2aPi inclusive. I visualize solutions to this sort of equation as angles in standard position that intersect the unit circle. (The angles can be graphed in the x-y plane with the vertex as the origin and the initial side along the positive x-axis). For x between 0 and 2Pi this amounts to making one complete "trip around the circle" and finding all possible solutions. Thus, each time a is increased by 1, there will be additional solutions to the equation -- "for every trip around the circle" there will be angles co-terminal to the solutions already found that are also solutions to the equation. I think you are being asked to generalize this for any integer value of a. The trickiest case is that for b = 0 since 1 additional "trip around the circle" will only yield 2 new solutions -- not 3.

Hope this helps.

for any sine,cosine of a constant number ( like 6, 7, 9, 3.7) we can use calculus.......we can use the Malcaurin series ...as follows....
example1)- what is the sin of (.4)?

ans cos(x)=1-x^2/2! + x^4/4!....and then we the the tangent can be expressed as tan=sin/cos=x-x^3/3!+x^5/5!-.../ 1-x^2!+x^4/4-...=x+1/3x^3=2/15x^5+...

hence, sin(x)=x-x^/3!+x^5/5!-x^7/7!+...(-1)^n-1 x^2n-1 /(2n-1)+...since this is a infinity series we let ... go on..
now since it goes on forever, i want to find the fifth degree of the malaurin series,....hence, P5 of x (not times x)= x-X63/3!+x^5/5!= ( since this is a limited function of the maclaurin series, we don't need a ...) f(x)=f(.4)=P5(.4)=.4-(.4)^3/3!+.4^5/5!=.378352

(note) ain(x)=x-x^3/3!+x^5/5!-x^7/7!+.....=sigma ^infinity n=0 (-1)^2 x^2x+1/(2n+1)! since the interval is infinity, sigma must have an infinity...