#$&* course MTH 164 Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: The cotangent function takes value -sqrt(3) when its argument is 5 pi / 6 or 11 pi / 6, as can easily be seen using a labeled unit circle and the definition of the cotangent. If the argument of the function is coterminal with 5 pi / 6 or 11 pi / 6 we obtain the same value for the cotangent. So the possible arguments are 5 pi / 6 + 2 pi n or 11 pi / 6 + 2 pi n, where n can be any integer. The student solution given below is correct for arguments 5 pi / 6 and 11 pi / 6, but does not address possible coterminal arguments. 2`theta/3 = 5pi/6 or 11pi/6 `theta = 5pi/6 * 3/2 or 11pi/6 * 3/2 `theta = 5pi/4 or 11pi/4
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23:43:44
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:2 **** List all the possible values of 2 `theta / 3 such that the equation is satisfied (the list is infinite; use the ellipsis ... to indicate the continuation of a pattern)
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2`theta/3 = 2/3(5pi/4 + kpi) when k=0 2/3`theta = 5pi/6 when k=1 2/3`theta = 11pi/6 when k=2 2/3`theta = 17pi/6 when k=3 2/3`theta = 235pi/6 when k=4 2/3`theta = 29pi/6 when k=5 2/3`theta = 35pi/6 when k=6 2/3`theta = 41pi/6 confidence rating #$&*: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): ok ------------------------------------------------ Self-critique Rating:2 **** How many of these values result in `theta values between 0 and 2 `pi?
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23:59:12 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your 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. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 2 `theta / 3 = 5 `pi / 6 + 2n `pi then multiplying both sides by 3/2 we get `theta = 5 `pi / 2 + 3 n `pi. Since 3 n `pi > 2 `pi only 5 `pi / 2 is between 0 and 2 `pi. If 2 `theta / 3 = 11 `pi / 6 + 2n `pi then multiplying both sides by 3/2 we get `theta = 11 `pi / 2 + 3 n `pi. Since 11 `pi / 2 > 2 `pi this can't lie between 0 and 2 `pi unless n is negative. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:2 **** Query problem 6.6.44 solve sin^2(`theta) = 2 cos(`theta) + 2
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14:41:12 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since sin^2(`theta) = 1 - cos^2(`theta) we have 1 - cos^2(`theta) = 2 cos(`theta) + 2. This equation is a quadratic equation in cos(`theta). To see this rearrange the equation to get cos^2(`theta) + 2 cos(`theta) + 1 = 0. Now let u = cos(`theta). You get u^2 + 2 u + 1 = 0. This is a quadratic equation with solution u = -1. My solution is u = -1, meaning cos(`theta) = -1, so `theta = -pi + 2 `pi k, for any integer k. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating:2 **** Query problem 6.6.66 19x + 8 cos(x) = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your 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) confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 19x+8cos(x)=2 by putting 19x+8 cos(x) in as y sub1 and 2 as y sub 2 we can trace and find that the closest value for x is approximately (.30).
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 2 " "Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution. Precalculus II Asst # 7 **** Query problem 6.6.12 cot(2`theta/3) = -`sqrt(3) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2`theta/3 = 5pi/6 or 11pi/6 `theta = 5pi/6 * 3/2 or 11pi/6 * 3/2 `theta = 5pi/4 or 11pi/4 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating:2 **** 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)
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your 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 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 2`theta/3 = 2/3(5pi/4 + kpi) when k=0 2/3`theta = 5pi/6 when k=1 2/3`theta = 11pi/6 when k=2 2/3`theta = 17pi/6 when k=3 2/3`theta = 235pi/6 when k=4 2/3`theta = 29pi/6 when k=5 2/3`theta = 35pi/6 when k=6 2/3`theta = 41pi/6 etc ........ when k=n 2/3`theta = 5pi/6 + npi ** 2 `theta / 3 can be 5 `pi / 6 or 5 `pi / 6 + 2 `pi or 5 `pi / 6 + 4 `pi or in general 5 `pi / 6 + 2 n `pi, or it can be 11 `pi / 6 or 11 `pi / 6 + 2 `pi or 11 `pi / 6 + 4 `pi or in general 11 `pi / 6 + 2 n `pi. **
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23:57:51
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:2 **** How many of these values result in `theta values between 0 and 2 `pi?
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23:59:12 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If 2 `theta / 3 = 5 `pi / 6 + 2n `pi then multiplying both sides by 3/2 we get `theta = 5 `pi / 2 + 3 n `pi. Since 3 n `pi > 2 `pi only 5 `pi / 2 is between 0 and 2 `pi. If 2 `theta / 3 = 11 `pi / 6 + 2n `pi then multiplying both sides by 3/2 we get `theta = 11 `pi / 2 + 3 n `pi. Since 11 `pi / 2 > 2 `pi this can't lie between 0 and 2 `pi unless n is negative. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique Rating:2 **** Query problem 6.6.44 solve sin^2(`theta) = 2 cos(`theta) + 2
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14:41:12 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your 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. My solution is u = -1, meaning cos(`theta) = -1, so `theta = -pi + 2 `pi k, for any integer k. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Since sin^2(`theta) = 1 - cos^2(`theta) we have 1 - cos^2(`theta) = 2 cos(`theta) + 2. This equation is a quadratic equation in cos(`theta). To see this rearrange the equation to get cos^2(`theta) + 2 cos(`theta) + 1 = 0. Now let u = cos(`theta). You get u^2 + 2 u + 1 = 0. This is a quadratic equation with solution u = -1. Thus our solution is u = -1, meaning cos(`theta) = -1, so `theta = -pi + 2 `pi k, for any integer k. **
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14:43:04
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:2 **** Query problem 6.6.66 19x + 8 cos(x) = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 19x+8cos(x)=2 by putting 19x+8 cos(x) in as y sub1 and 2 as y sub 2 we can trace and find that the closest value for x is approximately (.30) confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: 2 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: 2 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! 2 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!