#$&* course MTH 164 Precalculus II Asst # 7
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Given Solution: The cotangent function takes value –sqrt(3) when its argument is 5 pi / 6 or 11 pi / 6, as can easily be seen using a labeled unit circle and the definition of the cotangent. If the argument of the function is coterminal with 5 pi / 6 or 11 pi / 6 we obtain the same value for the cotangent. So the possible arguments are 5 pi / 6 + 2 pi n or 11 pi / 6 + 2 pi n, where n can be any integer. The student solution given below is correct for arguments 5 pi / 6 and 11 pi / 6, but does not address possible coterminal arguments. 2`theta/3 = 5pi/6 or 11pi/6 `theta = 5pi/6 * 3/2 or 11pi/6 * 3/2 `theta = 5pi/4 or 11pi/4 ......................................... 23:43:44 ......!!!!!!!!................................... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* Ok **** List all the possible values of 2 `theta / 3 such that the equation is satisfied (the list is infinite; use the ellipsis ... to indicate the continuation of a pattern) ......!!!!!!!!................................... YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2theta/3 = 2/3(5pi/4 + kpi) k value 0 5pi/6 1 11pi/6 2 17pi/6 3 235pi/6 4 29pi/6 5 35pi/6 this will continue on... confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 2`theta/3 = 2/3(5pi/4 + kpi) when k=0 2/3`theta = 5pi/6 when k=1 2/3`theta = 11pi/6 when k=2 2/3`theta = 17pi/6 when k=3 2/3`theta = 235pi/6 when k=4 2/3`theta = 29pi/6 when k=5 2/3`theta = 35pi/6 when k=6 2/3`theta = 41pi/6 etc ........ when k=n 2/3`theta = 5pi/6 + npi ** 2 `theta / 3 can be 5 `pi / 6 or 5 `pi / 6 + 2 `pi or 5 `pi / 6 + 4 `pi or in general 5 `pi / 6 + 2 n `pi, or it can be 11 `pi / 6 or 11 `pi / 6 + 2 `pi or 11 `pi / 6 + 4 `pi or in general 11 `pi / 6 + 2 n `pi. ** ......................................... 23:57:51 ......!!!!!!!!................................... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*Ok **** How many of these values result in `theta values between 0 and 2 `pi? ......!!!!!!!!................................... 23:59:12 Your solution: only 5pi/2 will lie between 0 and 2pi. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 2 `theta / 3 = 5 `pi / 6 + 2n `pi then multiplying both sides by 3/2 we get `theta = 5 `pi / 2 + 3 n `pi. Since 3 n `pi > 2 `pi only 5 `pi / 2 is between 0 and 2 `pi. If 2 `theta / 3 = 11 `pi / 6 + 2n `pi then multiplying both sides by 3/2 we get `theta = 11 `pi / 2 + 3 n `pi. Since 11 `pi / 2 > 2 `pi this can't lie between 0 and 2 `pi unless n is negative. ** ......................................... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*Ok **** Query problem 6.6.44 solve sin^2(`theta) = 2 cos(`theta) + 2 ......!!!!!!!!................................... 14:41:12 Your solution: sin^2(theta) = 2cos(theta) + 2 1-cos^2(theta) = 2cos(theta) + 2 cos^2(theta) + 2cos(theta) + 1 = 0 we can use the quadratic equation and find that cos(theta) = -1 not sure where to go from here. confidence rating #$&*1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Since sin^2(`theta) = 1 - cos^2(`theta) we have 1 - cos^2(`theta) = 2 cos(`theta) + 2. This equation is a quadratic equation in cos(`theta). To see this rearrange the equation to get cos^2(`theta) + 2 cos(`theta) + 1 = 0. Now let u = cos(`theta). You get u^2 + 2 u + 1 = 0. This is a quadratic equation with solution u = -1. Thus our solution is u = -1, meaning cos(`theta) = -1, so `theta = -pi + 2 `pi k, for any integer k. ** ......................................... 14:43:04 ......!!!!!!!!................................... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got alittle confused on this. ------------------------------------------------ Self-critique rating #$&*not bad **** Query problem 6.6.66 19x + 8 cos(x) = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 19x + 8cos(x) = 0 I’m really not sure where start on this one. confidence rating #$&*0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). ......................................... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I really wasn’t sure where to go on this one and when I looked at the given answer I am still confused because I do not know where the 2 came from in the given solution.