Solutions to Homework Problems from Section 7.5 of Stewart
The solutions of the basic equations cos x  k, sin x  k, and tan x  k are as follows:
1. If k is a number such that 1  k  1, then the solutions of the equation cos x  k are
x   arccosk  2n
(or  arccosk  360  n if we are using degrees).
2. If k is a number such that 1  k  1, then the solutions of the equation sin x  k are
x  arcsink  2n
and
x    arcsink  2n.
3. If k is any real number, then the solutions of the equation tan x  k are
x  arctank  n.
It is actually not a good idea to try to memorize the above three facts. It is better to rely on
your knowledge of the unit circle.
1. To solve the equation 2 cos x  1  0, we write this equation as
cos x  1
2
and observe that the solutions are
x     2n.
3
3. To solve the equation 2 sin x  3  0, we write this equation as
3
sin x 
2
and observe that the solutions are
x    2n
3
and
x  2  2n.
3
2
5. To solve 4 cos x  1  0, we write this as cos 2 x  1/4, and finally as cos x  1/2. The
terminal points on the unit circle that have first coordinate equal to 1/2 correspond to
x  /3, x  2/3. The solutions are thus
x     2n
3
and
x   2  2n.
3
7. If sec 2 x  2  0, then sec 2 x  2 which implies that cos 2 x  1/2 and that cos x   2 /2.
Using similar reasoning to that in problem 5, we obtain the solutions
x     2n
4
and
x   3  2n.
4
9. If cos x2 sin x  1  0, then either cos x  0 or sin x  1/2. The solutions of cos x  0
are
x     2n
2
and the solutions of sin x  1/2 are
x     2n
6
and
x  7  2n.
6
All of these are solutions of the original equation.
11. If tan x  3 cos x  2  0, then either tan x   3 or cos x  2, but the latter is
not possible because 2 is not in the range of the cosine function. Thus we must have
tan x   3 . Noting that arctan  3   /3, we obtain the solutions
x     n.
3
13. To solve cos x sin x  2 cos x  0, we write this equation as
cos xsin x  2  0
and note that all solutions of this equation must satisfy either cos x  0 (because it is not
possible that sin x  2). Therefore, the solutions are
x     2n.
2
2
15. The equation 4 cos x  4 cos x  1  0 can be written as
2 cos x  1 2  0.
Thus its solutions must satisfy 2 cos x  1  0 or cos x  1/2. The solutions of this
equation are thus
x     2n.
3
2
17. The equation sin x  2 sin x  3 can be written as sin 2 x  2 sin x  3  0 or as
sin x  3sin x  1  0.
The only solutions are those for which sin x  1 (because it is not possible that sin x  3).
The solutions are thus
x     2n.
2
2
2
19. The equation sin x  4  2 cos x can be written as sin 2 x  cos 2 x  4  2 cos 2 x  cos 2 x or
(using the Pythagorean identity) as
1  4  cos 2 x
or as
cos 2 x  3.
However it is clearly not possible that cos x   3 so this equation has no solutions. (The
graphs of y  sin 2 x and y  4  2 cos 2 x are shown below. Note that they do not intersect
anywhere.)
3
2
1
-4
-2
0
2 x
4
21. To solve 2 sin 3x  1  0, we write this equation as
sin 3x  1 .
2
This equation has solutions (let’s write them in degrees for a change)
3x  30   360  n
and
3x  150   360  n.
Thus the solutions are
x  10   120  n
and
x  50   120  n.
23. If cosx/2  1  0, then cosx/2  1 which means that
x  0   360  n
2
which means that
x  0   720  n.
25. To solve tan 5 x  9 tan x  0, we write this equation as
tan xtan 4 x  9  0
or, better yet, write it as
tan xtan 2 x  3tan 2 x  3  0.
There are thus three possibilities: solutions of tan x  0 which are
x  0   180  n,
solutions of tan x  3 , which (since arctan 3  60  ) are
x  60   180  n,
and solutions of tan x   3 which (since arctan  3
  60  ) are
x  60   180  n.
(Note that the equation tan 2 x  3  0 has no solutions. Why?)
27. To solve 4 sin x cos x  2 sin x  2 cos x  1  0, we write this equation as
2 sin x2 cos x  1  2 cos x  1  0
and then as
2 sin x  12 cos x  1  0.
The two possibilities are thus that sin x  1/2 and that cos x  1/2. The former equation
has solutions
x  30   360  n
and
x  150   360  n
and the latter equation has solutions
x  120   360  n.
29. To solve cos 2 2x  sin 2 2x  0, we write this equation as cos4x  0 (Recall the identity
cos2  cos 2   sin 2 .) The solutions of cos4x  0 are given by
4x  90   360  n.
Thus the solutions of this equation are
x  22. 5   90  n.
31. We can write 2 cos 3x  1 as cos 3x  1/2. We are only seeking solutions x such that
0  x  2 and these are obtained when 0  3x  6. The solutions are thus given by
3x   , 5 , 7 , 11 , 13 , 17
3 3 3
3
3
3
or
x   , 5 , 7 , 11 , 13 , 17 .
9 9 9
9
9
9
Here is a graph of y  2 cos 3x on the interval 0, 2. Note that it crosses the line y  1
six times.
2
1
0
1
2
3x
4
5
6
-1
-2
Graphs of y  2 cos3x and y  1
33. We want to find all solutions of 2 sin x tan x  tan x  1  2 sin x that lie in the interval
0, 2. Here are the graphs of y  2 sin x tan x  tan x and y  1  2 sin x on that interval:
3
2
1
0
1
2
3x
4
5
6
-1
-2
-3
It looks like the equation has four solutions. How can we find them? Let’s write the
equation as
tan x2 sin x  1  2 sin x  1
and then as
tan x2 sin x  1  2 sin x  1  0
and finally as
tan x  12 sin x  1  0.
From this we see that there are two possibilities: either tan x  1 or sin x  1/2. The
solutions of tan x  1 that lie in the interval 0, 2 are x  3/4 and x  7/4. The
solutions of sin x  1/2 that lie in the interval 0, 2 are x  /6 and x  5/6. These are
our four solutions! Note that
  0. 52360
6
3  2. 356 2
4
5  2. 618 0
6
7  5. 497 8
4
and this seems to agree with what we see in the above graph.
35. The equation tan x  3 cot x  0 can be written as tan x  3 cot x or as
tan x  3
tan x
or as
tan 2 x  3.
The solutions must thus satisfy tan 2 x   3 . The only such numbers x in the interval
0, 2 are
x   , 2 , 4 , 5 .
3 3 3 3
Are these all solutions? Let us graph the function y  tan x  3 cot x on the interval 0, 2:
3
2
1
0
1
2
3x
4
5
6
-1
-2
-3
Graph of y  tan x  3 cot x
It indeed appears to cross the x axis four times.
37. We would like to solve the equation tan 3x  1  sec 3x. First let’s graph y  tan 3x  1
and y  sec 3x on the interval 0, 2:
8
6
4
2
0
-2
1
2
3x
4
5
6
-4
-6
-8
It looks like there are many points of intersection. Can you count them? To solve the
equation
tan 3x  1  sec 3x
we square both sides to obtain
tan 3x  1 2  sec 2 3x
or
tan 2 3x  2 tan 3x  1  sec 2 3x.
Since tan 2 3x  1  sec 2 3x (a Pythagorean identity), we simply obtain
2 tan 3x  0
or
tan 3x  0.
The values of x that satisfy this equation for which 0  3x  6 are
3x  0, , 2, 3, 4, 5.
The possible solutions for which 0  x  2 are thus
x  0,  , 2 , , 4 , 5 .
3 3
3
3
Note that I said possible solutions. The only ones that actually work when we plug them
back into the original equation are x  0, x  2/3, and x  4/3. For example, if we try
x  , we obtain
tan3  1  0  1  1
but
1
 1  1
sec3 
1
cos3
which shows that x   is not a solution. Why did we find some extra solutions? The
answer is that this is always a possibility when we attempt to solve an equation by
squaring both sides. Now here is the really surprising part: When we first looked at the
graphs, it looked like they had many points of intersection. However, there are only three!
Since the graphs lie very close to each other at certain places, it just looks like they
intersect but they really don’t. This is an example of why we can’t always rely on
graphical analysis. Looking at the graphs again, do you now see the three points of
intersection? (Note that x  2 is actually a fourth point of intersection but we are not
counting it since we were asked only to find solutions in the interval 0, 2.)
59. The equation sin x  sin 3x  0 can be written as
2 sin2x cosx  0.
The solutions must thus satisfy either sin2x  0 or cosx  0. The solutions of
sin2x  0 are
2x  0   360  n
and
2x  180   360  n
or
x  0   180  n
and
x  90   180  n.
The solutions of cosx  0 are
x  90   360  n.
61. Since cos4x  cos2x  2 cos3x cosx, the equation cos4x  cos2x  cosx can
be written as
2 cos3x cosx  cosx
or as
2 cos3x cosx  cosx  0
or as
cosx2 cos3x  1  0.
The solutions of cosx  0 are given by
x  90   360  n
and the solutions of cos3x  1/2 are given by
3x  60   360  n
or
x  20   120  n.
Thus all solutions of the equation cos4x  cos2x  cosx are given by
x  90   360  n
and
x  20   120  n.
Here are graphs of y  cos4x  cos2x and y  cosx on the interval 0  , 360  :
2
1.5
1
0.5
0
50
100
150
x 200
250
300
-0.5
-1
The eight points of intersection that we see in this picture are
90   360   1  270 
90   360   0  90 
20   120   1  100 
20   120   2  220 
20   120   3  340 
20   120   0  20 
20   120   1  140 
20   120   2  260 
350