Solve a sin x + b cos x = c
Yue Kwok Choy
Question
Solve :
12 sin x + 5 cos x = 4
P
Method 1
Construct a triangle OPQ with

tan  
12
.
5
13
12
  67.380o .

Also,
OP  5 2  12 2  13
Now,
12 sin x + 5 cos x = 4
O
Q
5
(13 sin ) sin x + (13 cos ) cos x = 4
13 (cos x cos  + sin x sin ) = 4
13 cos (x – ) = 4
cosx    
4
13
x  67.380   360  n  72.080 o
x  360  n  72.080 o  67.380 

x  360  n  139.460 
x  360  n  4.700  ,where n
or
is an integer . …. (1)
Method 2
Construct a triangle OPQ with

tan  
5
.
12
P
  22.620o .
13
Also,
OP  5 2  12 2  13
Now,
12 sin x + 5 cos x = 4
5

(13 cos ) sin x + (13 sin ) cos x = 4
O
12
Q
13 (sin x cos  + cos x sin ) = 4
13 sin ( x + ) = 4
sin x   
4
13
x  22.620   180  n   1 17.920 o
n
x  180  n   1 17.920 o  22.620  , where n is an integer .
n
…. (2)
To show that the solution set (2) is the same as the solution set (1), we need to divide the solution into
odd and even cases :
(a)
n = 2m ,
(b) n = 2m + 1,
x  180  2m  17.920 o  22.620   360  m  4.700 
x  180  2m  1  17.920 o  22.620   360  m  139.460  , where m is an integer.
1
(t – method)
Method 3
t  tan
Let
x
.
2
x
x
2 tan
x
2 cos 2 
2  2t
x
x 1  t2
2
cos
sec2
2
2
sin
x
x
sin x  2 sin cos  2
2
2
x
x

sin 2 
1  tan 2

2
x
x
x
2  cos 2 
2  1 t
cos x  cos 2  sin 2  1 
x
x
2
2 
2
1  t2
cos 2 
sec2

2
2

Now, the equation :
12 sin x + 5 cos x = 4
2
 2t   1  t 

4
12

5

2
2 

1  t  1  t 

 
122t   5 1  t 2  4 1  t 2

9t  24t  1  0
2
t
4  17
3

tan
x
 2.7077 or
2
tan

x
 180  n  69.730 
2
or
x
 180  n  2.3500 
2

x  360  n  139.460 
or
x  360  n  4.700  , where n is an integer .
Method 4
x
 0.041035
2
…. (3)
(unsatisfactory)
12 sin x + 5 cos x = 4
12 sin x  5 1  sin 2 x  4
 5 1  sin 2 x  4  12 sin x


25 1  sin 2 x  16  96 sin x  144 sin 2 x
169 sin 2 x  96 sin x  9  0
sin x 
48  15 17

169
0.64998
x  180  n   1 40.540 
n
or
or
0.081932

x  180  n   1  4.700 
n

Method 4 is unsatisfactory since solution set (4) gives a larger solution set than
Readers please explain. What solutions should be rejected ?
…. (4)
(1), (2) or (3).
How to reject ?
2
Explanation
Squaring equation creates roots!
That is why Method 4 is not recommended.
Finding the redundant roots is tedious:
In (4),
(a)
When
(i)
x  180  n   1 40.540  .
n
x  360 m  40.540
n = 2m,
x  360  m  139.460 
(ii) n = 2m + 1,
(b) When
(i)
n

x  360 m  4.700 

n = 2m,
x  360  m  184.700 
(ii) n = 2m + 1,
Therefore

x  180  n   1  4.700 
(4) is equivalent to :
 x  360  m  40.540 



x  360 m  139 .460



 x  360 m  4.700
x  360  m  184 .700 

....(4.1)
....(4.2)
....(4.3)
....(4.4)
(4.2) and (4.3) are good solutions.
(4.1) and (4.4) should be rejected.
For (4.1), 12 sin x + 5 cos x



 12 sin 360  m  40.540   5 cos 360  m  40.540 

 12 sin 40.540   5 cos 40.540 
 120.64998   50.75995 
 11.59951
4

x  360  m  40.540 
Similarly, for (4.4),
does not satisfy the equation
12 sin x + 5 cos x = 4, and should be rejected.
x  360  m  184.700  should be rejected.
3