SOLUTION: Given the equation sin x = −0.2, find only the solutions that are in the interval [−π, 5π]. Round to nearest 0.01. #confused. Im plugged that into m

Algebra ->  Trigonometry-basics -> SOLUTION: Given the equation sin x = −0.2, find only the solutions that are in the interval [−π, 5π]. Round to nearest 0.01. #confused. Im plugged that into m      Log On


   

Question 909017: Given the equation sin x = −0.2, find only the solutions that are in the interval [−π, 5π].
Round to nearest 0.01.

#confused.
Im plugged that into my calculator and gotten. x=-.20 in radians. However the interval part of the question is throwing me off.
Found 2 solutions by Edwin McCravy, Theo:
Answer by Edwin McCravy(20081) About Me  (Show Source):
You can put this solution on YOUR website!
You can't go by what the calculator gives for sin-1 unless
you want only the answer between -pi%2F2 and %22%22+%2B+pi%2F2.

Also it's better not to round off until the end.

What you do is fisrt find sin-1 of POSITIVE 0.2 which will
give the QI answer of .2013579209, which is not a solution, but only
the reference angle in radians.  

First let's find the positive values, then we'll find the negative values.

We know that the sine is negative only in QIII and QIV,  So to get the 
first positive QIII answer we add pi to the reference angle to get
3.342950574.  To get the first QIV answer we subtract the reference angle
from 2pi and get 6.081827386.

So the first two positive solutions are 3.342950574 and 6.081827386

Now we begin adding 2pi to each of those and get as many answers
as we can that don't exceed 5pi which is 15.70796327.

Adding 2pi to 3.342950574 and 6.081827386 gives

                    9.626135882 and 12.36501269

Adding 2pi to those gives

                   15.90932119 and 18.648198, but both are too large.

So all the positive solutions are

3.342950574, 6.081827386, 9.626135882 and 12.36501269.

Now let's find the negative solutions.

We go bact to the first two positive solutions, 3.342950574 and 6.081827386

Now we begin subtracting 2pi from each of those and get as many 
negative solutions as we can that don't go below -pi which is 
-3.141592654.

Subtracting 2pi from 3.342950574 and 6.081827386 gives
                        -2.940234733 and -.2013579208

Subtracting 2pi from those gives

                       -9.22342004 and -6.484543228 both too small.

So all the solutions are, smallest to largest:

       -2.94, -0.20, 3.34, 6.08, 9.63, 12.37 
                           
Edwin

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
there are 4 quadrants in the interval from 0 to 2pi.
first quadrant is 0 to pi/2 (same as 0 to 90 degrees)
second quadrant is pi/2 to pi (same as 90 degrees to 180 degrees)
third quadrant is pi to 3pi/2 (same as 180 to 270 degrees)
fourth quadrant is 2pi(same as 270 to 360 degrees)

once you get past 2pi, the cycle repeats itself.

your solutions will be in the interval of 0 to 2pi and then they will repeat themselves.

let's do you problem and you'll see what i mean.

if you use your calculator to find the angle whose sine is -.2, the calculator will give you an angle of -.2013579208 radians.

that angle is in the fourth quadrant.

that doesn't really help you too much.

it's better to find the angle in the first quadrant.

you do that by making the sine positive since all trig functions are positive in the first quadrant.

when you solve for sinx = .2, you get an angle of .2019579208 radians.

that becomes your reference angle.

you can use that angle to find all the other angles.

now you know the sine is negative because that what you were given.

the sine is negative in quadrants 3 and 4.

the rules for finding an angle in each quadrant are as follows:

your reference angle is in quadrant 1.

the rules for finding the angle in each quadrant are as follows:

the angle in quadrant 1 is the reference angle.
the angle in quadrant 2 is pi minus the reference angle.
the angle in quadrant 3 is pi plus the reference angle.
the angle in quadrant 4 is 2pi minus the reference angle.

you want the angle in quadrant 3 and quadrant 4.

your angle in quadrant 1 is .2019579208 radians which is your reference angle.

your angle in quadrant 3 will be pi + .2019579208 radians = 3.342950574 radians

your angle in quadrant 4 will be 2pi - .2019579208 = 6.081827386 radians

your angle will be multiples of plus or minus 2pi radians until you get past the interval you are interested in.

the interval you are interested in is -pi radians to 5pi radians.

-pi radians is equal to -3.142592654 radians.

5pi radians is equal to 15.70796327 radians.

your interval of interest is from -3.142592654 radians to 15.70796327 radians.

take your angle in quadrant 2 and keep adding 2*pi until you get past 15.70796327 radians

you will get:

3.342950574 + 1*2pi = 9.626135882 radians
3.342950574 + 2*2pi = 15.90932119 radians ***** > 15.70796327 radians

take your angle in quadrant 2 and keep subtracting 2*pi radians until you get past -3.142592654 radians.

3.342950574 - 1*2pi = -2.940234733 radians.
3.342950574 - 2*2pi = -9.22342004 radians ***** < -3.142592654 radians

take your angle in quadrant 4 and do the same.

6.081827386 + 1*2pi = 12.36501269 radians
6.081827386 + 2*2pi = 18.648198 radians ***** > 15.70796327 radians

6.081827386 - 1*2pi = -.2013579208 radians
6.081827386 - 2*2pi = -6.484543228 radians ***** < -3.142592654 radians

your solutions are:

-2.940234733 radians
-.2013579208 radians
3.342950574 radians
6.081827386 radians
9.626135882 radians
12.36501269 radians

this will be easier to see on a graph.
any difference in the numbers you see on the graph and the numbers shown above are due to rounding.

your graph will look like this:

$$$