Given that y is measured in radians, find the two smallest positive values of y such that
6sin(2y + 1) + 5 = 0
How?
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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.Quote:
Originally Posted by Yunie
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.