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MATH 1730
Pre-Calculus
Fall 2014
Dwiggins
Homework Assignment # 7
4.9. Trigonometric Equations.
page 262, # 20. As demonstrated in class, this equation is solved by factoring:
2sin2x – 3sinx + 1 = (2sinx – 1)(sinx – 1) = 0
sinx = ½ or sinx = 1, which means the unit circle solution set is given by
{/6, /2. 5/6}. (Note we were told x is a real number, so give the answers
in radians, not degrees. Also, to find all possible solutions,
shift any of the unit circle solutions by any multiple of 2.
# 26.
2sin2 + 2sin – 3sin – 3 = (2sin – 3)(sin + 1) = 0
sin = 3/2 or sin = –1
  is either in the set {60, 150, 270) or has one
of these values shifted by a multiple of 360
# 38. Using the unit circle identity sec2 = tan2 + 1, we have
tan4 – 2sec2 + 3 = tan4 – 2tan2 + 1 = (tan2 – 1)2 = 0
 tan2 =1 tan = 1
  = 45, 135, 225, 
# 50. Using the sum-to-product rule found on page 244 of the text, we have
sin5 + sin3 = 2sin4cos = 0
 sin4 = 0 or cos = 0,
and so the values of  in the interval [0,2) which give sin5 + sin3 = 0
are 0, /4, /2, 3/4, , 5/4, 3/2, and 7/4.
page 263, # 64. The function f(x) = cosx + x + 1 is equal to zero when the graph of y = cosx
crosses the straight line y = –x – 1. When these two graphs are plotted, it is
evident that the line crosses the cosine curve at precisely one point, somewhere
between x = –/2 and x = –1. In fact, since f(–/2) = 0 –/2 + 1 < 0 and
f(–1) = cos(–1) + 0 = cos(1) > 0, we must have f(x) = 0 somewhere in between.
Using a little bit of first-semester calculus, I find f(x) = 0 at approximately x = –1.28.
page 263, # 64. 3sin2x – 8sinx + 4 = (3sinx – 2)(sinx – 2) = 0
sinx = 2/3 or sinx = 2, and the latter case is not possible, since |sinx| < 1.
Also, we are told |x| < /2, and so we take the acute value of x = sin–1(2/3) = 0.73,
which corresponds to an angle of measure 41 50.
# 74.
I(t) = 30sin[120(t – 7/36)] = 30sin[120t – 70/3 + 24] = 30sin[120t + 2/3],
and so I = 15 when sin[120t + 2/3] = ½, which occurs when 120t + 2/3 = /6,
but this would give a negative value of t. Thus, we choose the next value of  for which
sin  = ½, namely  = 5/6, and then set 120t + 2/3 = 5/6 120t = /6 to obtain
t = 1/720 (= 0.0014 second) as the first positive value of t which gives I = 15.
5.1. Solving Right Triangles.
page 283, # 12.
c = 10,  = 49 ,  = 90 – 49 = 41
a = 10 cos 49 = 6.56 , b = 10 sin 49 = 7.55 .
# 18.
b = 4,  = 58 ,  = 90 – 58 = 32
a = 4 tan 58 = 6.40 , c = 4 sec 58 = 7.55
# 20. c = 6, b = 3 = c/2 . 30-60-90 triangle . a = 33 ,  = 60 ,  = 30
# 24.
If we let a denote the length of the base in the 46 triangle, then using tangent ratios
gives both x = a tan 46 and x = (a + 10) tan 34 = atan34 + 10tan34
a(tan 46tan3410tan34a = 18.68x = 19.35
page 287, # 2. dist(T1, T2) = (100 ft)tan(29.7) = 49.55 ft = 49 ft 6.6 inch
page 288, # 14.
a = 1000tan(79.946) = 5640.19, b = 1000tan(80.05) = 5700.37
h = b – a = 60.18 ft = 60 ft 2.2 inch