Question 170144
1) If x is an angle in quadrant 4 and cotx= -7/24, find the value of 
sin[(1/2)x].
If cotx = -7/24, x = 7 and y = -24
Then r = sqrt(7^2 + 24^2) = 25
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sin[(1/2)x] = sqrt[(1-cos(x))/2]
So, sin[(1/2)x] = sqrt[(1-(7/25))/2] = sqrt[9/25] = 3/5
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2) What is the solution set of the equation sin(2x)-cos^2(x+1) = sin^2(x)+sinx  in the interval 0<x<2pi?
Comment: That is a mess to analyze.  Graph the left side and the right side
separagely and see where they intersect under the condition the 0<x<2pi.
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3) Find the solution set of 6sinx + 11= 2cscx over the domain 0<x<360.
Multiply thru by sin(x) to get:
6sin^2(x) + 11sin(x) - 2 = 0
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Let w = sin(x)
Substitute to get:
6w^2 + 11w - 2 = 0
6w^2 +12w - w -2= 0
6w(w+2) -(w+2) = 0
(w+2)(6w-1) = 0
w = -2 or w = 1/6
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Solve for x:
sin(x) = -2 or sin(x)= 1/6
sin(x) cannot be -2.
If sin(x) = 1/6, x = 9.594.. degrees or 170.41 degrees
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Cheers,
Stan H.