Find all values of x in the interval (0, 2pie) that satisfy the equation sin(x) = sin(2x)?

Question

please help me i have a quiz on this kind of question in 45 min and i am really worried about this question. the formula of sin(x) = sinc (2x) is sin(2x) = cos2x-sinn2x

Answer 1

first of all, the transcendental number you've cited is spelled 'pi,' not 'pie.'

(0, 2pi) is a classic interval because it includes every angle through a circle once; you can think of it as (0, 360) in degrees.

so, you have this equation,

six(x) = sin(2x)

so either

x = 2x,

in which case x = 0 (which is one of the answers), or

2x = x + 2pi, or
x = 2pi (which is another of the answers)

however, there is one answer left. the only case in which the value of sin(x) repeats more often than every 2pi is when

sin(x) = 0, or
x = (n)pi

including (1)pi. (this is the third solution.)

so, x = {0, pi, 2pi}.

Answer 2

Hello

Sin(x) = sin(2) at 0, pi/2, and pi.

But since you are not including these in your interval (0, 2pie) - the only answer is pi/2.

Hope this helps

Answer 3

sin x =2 sin x cos x so
sin x ( 1-2cos x ) = 0
sin x = 0 so x= 0 and pi
1-2cos x = 0 cos x= 1/2 and x= pi/3 and x= 5pi/3