The discussion revolves around solving trigonometric equations within the range of 0 to 360 degrees. The specific equations under consideration are 2sin x + sin x cos x = 0 and sin 2x = -1/sqrt(2).
Participants are actively engaging with the problems, with some offering partial solutions and others questioning the validity of certain steps. There is an acknowledgment of the need for further exploration of the second equation, particularly regarding the values of 2x.
Some participants express uncertainty about their previous attempts and the correctness of their methods. There is a mention of a lack of familiarity with LaTeX, which may affect the clarity of future contributions.
Show us what you did and we'll go from there...xdeanna said:I had some questions on a test and i got these two wrong..
x is greater than or equal to 0 and x is less than or equal to 360
1) 2sin x + sin x cos x = 0
2) sin 2x = -1/sqrt(2)
This is one solutions; there are two other numbers x in [0, 360] (deg.) for which sin(x) = 0. The inverse sine function will only produce numbers in the interval [-90, 90], in degrees. Think about where the graph of y = sin(x) crosses the x-axis.xdeanna said:2Sin x + cos x sin x = 0
Sin x(2 + cos x) = 0
sin x = 0
x= sin^-1(0)
x=0degrees
xdeanna said:or
2+ cos x= 0
cos x =-2
x= cos^-1(-2)
x=?
I did something different on the test but i think this is wrong too >.<
Don't make show what I did for #2 :(