Solve the equation?

Question

Solve the equation sinx = sin(x+2) for x is between or equal to 0 and 2 pie. (hint: Use the unit circle definition of sine for circular funtions)

Answer: (pie/2 -1, 3pie/2 -1)

Answer: (pie/2 -1, 3pie/2 -1)

Answer 1

Bah.  I'm amazed that nobody can post an analytic solution.

sin(x) is symmetrical around points of (n+1/2)π, where n is integer.  Ergo, when x=((n+1/2)π-1), sin(x+2) = sin(x).

Answer 2

just use the additiona formula!

sin(x) = sin(x+2)
= sin(x)sin(2) + cos(x)cos(2), gather sines on left
sin(x)-sin(x)sin()=cos(x)cos(2); factor left side
sin(x)[1-sin(2)] = cos(x)cos(2); divide both sides by cos(x)

sin(x)/cos(x) = cos(2)/[1-sin(2)], so
x = arctan(cos(2)/[1-sin(2)]

Look this up in a table or a scientific calculator.

Answer 3

sinx = sin(x + 2)
sin (x + 2) = sin x cos 2 + cos x sin 2
cos(2) + sin(2)cot(x) = 1
sin(2)cot(x) = 1 - cos(2)
cot(x) = (1 - cos(2))/sin(2)
tan(x) = sin(2)/(1 - cos(2)) ≈ 0.6420926
x = 0.5707963, 3.7123890

Answer 4

sin(x+2) = sin(x)cos(2)+sin(2)cos(x)

sin(x) = sin(x)cos(2) + sin(2)cos(x)
1 = cos(2) + sin(2)/tan(x)
(1-cos(2))/sin(2) = tan(x)
x = arctan((1-cos(2))/sin(2))

Answer 5

sin(x) never equals sin(x+2) as I am aware of it.

Maybe you've made a typo?