Find solutions of sin x=1/2 for the interbal (o, 2pie). What happens if th einterval changes to (negative pie, pie)? (0, 4pie), (0, 8pie)?

I think i got the answer but i'm not quite sure. please help! Thanks

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- Mar 9th 2006, 02:21 PMaussiekid90Trigonometry
Find solutions of sin x=1/2 for the interbal (o, 2pie). What happens if th einterval changes to (negative pie, pie)? (0, 4pie), (0, 8pie)?

I think i got the answer but i'm not quite sure. please help! Thanks - Mar 9th 2006, 03:22 PMJameson
So start with the reference angle. sin(x)=1/2 at $\displaystyle \frac{\pi}{6}$. Now go into quadrant two, add $\displaystyle \frac{\pi}{2}$ and the sine graph is still positive. This means that sin(x)=1/2 at $\displaystyle \frac{2\pi}{3}$ as well. In the 3rd and 4th quadrants the sine graph is negative, so you just have two answers.

- Mar 10th 2006, 01:46 AMticbolQuote:

Originally Posted by**aussiekid90**

Positive sine, so angle X is in the 1st or 2nd quadrants.

X = arcsin(1/2) = 30deg = pi/6 rad.

For the interval (0,2pi):

X = pi/6 ---in the 1st quadrant, or,

X = pi -pi/6 = 5pi/6 ---in the 2nd quadrant

Therefore, X = pi/6 or 5pi/6, in radians. ------------answer.

For the interval (-pi,pi):

So the interval starts at -pi, goes counterclockwise, passes -pi/2, passes 0, passes pi/2, and ends at pi.

When laid out in a straight line, like in the horizontal axis of a graph, the interval starts at -pi, going to the right, passes -pi/2, passes 0, passes pi/2, and ends at pi.

The -pi is the divider of the 3rd and 2nd quadrants. Likewise, the pi is the divider of 2nd and 3rd quadrants. Hence, the start and finish of the interval is the negative side of the horizontal axis.

[In the (0,2pi) interval, the positive side of the horizontal axis is the start and finish of the interval. But, in both intervals, since (...) were used, the -pi and pi, or the 0 and 2pi, are not included in their respective intervals.]

So, since the angle X is in the 1st and 2nd quadrants, then

X = pi/6 or 5pi/6 radians still. ------------answer.

For the interval (0,4pi):

X is still in the 1st or 2nd quadrants.

In the first revolution, from 0 to 2pi, X = pi/6 or 5pi/6.

In the second revolution, from 2pi to 4pi, X = (pi/6 +2pi) = 13pi/6, or (5pi/6 +2pi) = 17pi/6.

Therefore,

X = pi/6, 5pi/6, 13pi/6, or 17pi/6 radians. ---------answer.

Check if 17pi/6 is correct,

sin(17pi/6) = 0.5, so, Ok.

For the interval (0,8pi):

The 8pi is 4 revolutions, so,

X = pi/6, (pi/6 +2pi), (pi/6 +4pi), (pi/6 +6pi) in the 1st quadrant.

X = 5pi/6, (5pi/6 +2pi), (5pi/6 +4pi), (5pi/6 +6pi) in the 2nd quadrant.

For your exercise, try to simplify those.

You should get

X = pi/6, 5pi/6, 13pi/6, 17pi/6, 25pi/6, 29pi/6, 37pi/6, or 41pi/6 radians -----answer.