#$&* course MTH 164 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: The cotangent function takes value -sqrt(3) when its argument is 5 pi / 6 or 11 pi / 6, as can easily be seen using a labeled unit circle and the definition of the cotangent. If the argument of the function is coterminal with 5 pi / 6 or 11 pi / 6 we obtain the same value for the cotangent. So the possible arguments are 5 pi / 6 + 2 pi n or 11 pi / 6 + 2 pi n, where n can be any integer. The student solution given below is correct for arguments 5 pi / 6 and 11 pi / 6, but does not address possible coterminal arguments. 2`theta/3 = 5pi/6 or 11pi/6 `theta = 5pi/6 * 3/2 or 11pi/6 * 3/2 `theta = 5pi/4 or 11pi/4
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23:43:44
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wasn’t quite sure about this one. I just didn’t understand. ------------------------------------------------ Self-critique Rating:
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5pi/6 11pi/6 17pi/6 235pi/6 29pi/6 35pi/6 41pi/6 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 2`theta/3 = 2/3(5pi/4 + kpi) when k=0 2/3`theta = 5pi/6 when k=1 2/3`theta = 11pi/6 when k=2 2/3`theta = 17pi/6 when k=3 2/3`theta = 235pi/6 when k=4 2/3`theta = 29pi/6 when k=5 2/3`theta = 35pi/6 when k=6 2/3`theta = 41pi/6 etc ........ when k=n 2/3`theta = 5pi/6 + npi ** 2 `theta / 3 can be 5 `pi / 6 or 5 `pi / 6 + 2 `pi or 5 `pi / 6 + 4 `pi or in general 5 `pi / 6 + 2 n `pi, or it can be 11 `pi / 6 or 11 `pi / 6 + 2 `pi or 11 `pi / 6 + 4 `pi or in general 11 `pi / 6 + 2 n `pi. **
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23:57:51
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** How many of these values result in `theta values between 0 and 2 `pi?
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23:59:12 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 2 `theta / 3 = 5 `pi / 6 + 2n `pi then multiplying both sides by 3/2 we get `theta = 5 `pi / 2 + 3 n `pi. Since 3 n `pi > 2 `pi only 5 `pi / 2 is between 0 and 2 `pi. If 2 `theta / 3 = 11 `pi / 6 + 2n `pi then multiplying both sides by 3/2 we get `theta = 11 `pi / 2 + 3 n `pi. Since 11 `pi / 2 > 2 `pi this can't lie between 0 and 2 `pi unless n is negative. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query problem 6.6.44 solve sin^2(`theta) = 2 cos(`theta) + 2
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14:41:12 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `theta = -pi + 2 `pi k confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Since sin^2(`theta) = 1 - cos^2(`theta) we have 1 - cos^2(`theta) = 2 cos(`theta) + 2. This equation is a quadratic equation in cos(`theta). To see this rearrange the equation to get cos^2(`theta) + 2 cos(`theta) + 1 = 0. Now let u = cos(`theta). You get u^2 + 2 u + 1 = 0. This is a quadratic equation with solution u = -1. Thus our solution is u = -1, meaning cos(`theta) = -1, so `theta = -pi + 2 `pi k, for any integer k. **
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14:43:04
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query problem 6.6.66 19x + 8 cos(x) = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 0.31 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 19x+8cos(x)=2 by putting 19x+8 cos(x) in as y sub1 and 2 as y sub 2 we can trace and find that the closest value for x is approximately (.30).
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!