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this was one of the problems i need help with. Thanks.     solve for x in the given interval. You should be able to find these numbers using reference triangles in the proper quadrants.

csc x=2      (pie)/2 is less than equal to x is less than equal to (pie)

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Questioner:   stacey
Category:  Advanced Math
Private:  No
 
Subject:  Pre-Calc
Question:  this was one of the problems i need help with. Thanks.     solve for x in the given interval. You should be able to find these numbers using reference triangles in the proper quadrants.

csc x=2       pi/2 <= x <= pi
......................................................
Hi, Stacey,

[See abbreviated notation above for future ease in typing.]

The restriction:  pi/2 <= x <= pi
means that you must find a solution that is in the second quadrant.  You may think of it as 90 degrees to 180 degrees, but don't write that.  Real men don't use degrees, they use radians. (Real women, too.)

If  csc @ = 2, then  sin @ = 1/2.  Most people feel uncomfortable with values of secant or cosecant, so we change to sine or cosine, using the reciprocal identities:

csc @ = 1/sin @ and vice versa.
sec @ = 1/cos @ and vice versa.

I have changed the variable to @ [theta], because we use x and y for other things in this context.

Since the definition of sin @ = y/r, you will assign  y = 1, r = 2, and then draw the appropriate

reference triangle.

That's difficult in this klutzy interface, but I'll do my best:

********** VIEW IN COURIER FONT ************

          y-axis
    |\        |
    | \       |
    |  \r=2   |
    |   \     |
y=1 |    \    |
    |     \   |
    |      \  |
    |       \ |
    |        \|
-----+---------+------
              |
[That's not drawn to scale -- it's the best I can do.]

From the Pythagorean theorem,  x = - sqrt(3)/2.  [Minus, because this is in the second quadrant.]

Now use your knowledge of the special triangles.  The angle that is IN THE TRIANGLE and AT THE ORIGIN, called the REFERENCE ANGLE, is  30 degrees.  [Ok,OK, it's pi/6]

That means the value of @ in this position, second quadrant, is 150 degrees or 5pi/6.

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