trigonometry

[Yunie]

Given that y is measured in radians, find the two smallest positive values of y such that

6sin(2y + 1) + 5 = 0

How?

[krtxmrtz]

Given that y is measured in radians, find the two smallest positive values of y such that

6sin(2y + 1) + 5 = 0

How?

Hint: sin(x) is a periodic function and is 0 for x = k*Pi, where k is any integer.

[03myersd]

Lets see if I can remember how to do this:

6Sin(2y+1) + 5 = 0
6Sin(2y+1) = -5
Sin(2y+1) = -5/6

So Sin is negative. Which means that the results must lie in the 3rd and 4th quadrant.

1st: Sin-1(5/6) = 56.4426902380793
3rd: 180 + 56.4426902380793 = 236.442690238079
4th: 360 - 56.4426902380793 = 303.557309761921

2y + 1 = 236.443, 303.557
2y = 235.443, 302.557
Y = 117.722,151.279

In degress y = 115.722 and y = 151.279
In radians y = 2.020 and y = 2.640

This was assuming that 0 <= y <= 2*pi but its the two smallest values anyway so it doesn't really matter.

I apologise if this is wrong.