Trigonometric Equations 2
Lecture Notes
page 1
Sample Problems
Solve each of the following equations.
1.) sin x =
sin2 x
6.) cos x sin x = cos x
2.) 2 cos2 x
5 cos x = 3
7.) tan x = tan2 x
3.) 3 (1
sin x) = 2 cos2 x
8.) 2 cos x
sin x + 2 cos x sin x = 1
4.) sin x =
cos x
9.) cos x = 1 + sin2 x
5.) tan2 x =
1
3
10.) 7 sin x + 5 = 2 cos2 x
Practice Problems
Solve each of the following equations.
1.) 1 + sin x = 2 cos2 x
2.)
8.) tan x sin2 x =
3 cos x + 3 = 2 sin2 x
3.) cos3 x = cos2 x
4.) 2 cos2 x
cos x = 3
9.)
3
tan x
4
cos x
1 + sin x
+
=4
cos x
1 + sin x
10.) sin2 x
5.) 2 sin2 x = cos x + 1
1
1
cos x + cos x sin2 x =
2
2
11.) cot x = cos x
6.) 2 cos2 x + 3 sin x = 3
2
7.) sec x = 4
c copyright Hidegkuti, Powell, 2009
12.) tan x
p
2=
1
p
tan x + 2
Last revised: May 26, 2010
Trigonometric Equations 2
Lecture Notes
page 2
Sample Problems - Answers
1.) x = k1 and x =
3.) x =
2
4.) x =
7.) x =
+ 2k1 and x =
4
4
8.) x =
2
+k
6
+ 2k2 and x =
where k 2 Z
+ k1 and x = k2
3
10.) x =
+ 2k1 and x =
5
+ 2k3
6
5.) x =
6
2
+ 2k
3
2) x =
where k1 ; k2 ; k3 2 Z
where k 2 Z
+k
where k 2 Z
6.) x =
2
+k
where k 2 Z
where k1 ; k2 2 Z
+ 2k1 and x =
6
where k1 ; k2 2 Z
+ 2k2
+ 2k2
where k1 ; k2 2 Z
7
+ 2k2
6
where k1 ; k2 2 Z
2
9.) x = 2k
where k 2 Z
Practice Problems - Answers
1.) x =
6
2.) x =
+ 2k1 and x =
3
+ 2k1 and x = 2k2
4.) x =
+ 2k
6.) x =
+ 2k1 ; x =
2
7.) x =
9.) x =
11.) x =
3
3
2
5
3
+ 2k2 and x =
+ 2k3 where k1 ; k2 ; k3 2 Z
6
2
where k 2 Z
6
+k
5.) x =
+ 2k2 ; and x =
+ k where k 2 Z
+ 2k
where k1 ; k2 2 Z
where k 2 Z
where k 2 Z
c copyright Hidegkuti, Powell, 2009
3
3.) x =
and
+ 2k1
5
+ 2k3
6
8.) x = k ; x =
10.) x =
12.) x =
4
3
2
+ k1 ; and x = 2k2 ; where k1 ; k2 2 Z
where k1 ; k2 2 Z
+ 2k2
where k1 ; k2 ; k3 2 Z
3
+ 2k ; and x =
+ k1 ; and x =
+k
2
+ 2k
3
+ 2k2
where k1 ; k2 ; k3 2 Z
where k1 ; k2 2 Z
where k 2 Z
Last revised: May 26, 2010
Trigonometric Equations 2
Lecture Notes
page 3
Sample Problems - Solutions
sin2 x
1.) sin x =
Solution:
sin x =
sin2 x
sin2 x + sin x = 0
sin x (sin x + 1) = 0
sin x = 0
or
add sin2 x
factor out sin x
sin x + 1 = 0
sin x = 1
If sin x = 0; then x = k where k 2 Z: If sin x + 1 = 0; then sin x =
2.) 2 cos2 x
Solution:
1 and so x =
2
+ 2k where k 2 Z.
5 cos x = 3
This equaton is quadratic in cos x: If helps, we may introduce a new variable, a = cos x:
2 cos2 x
Let a = cos x
5 cos x = 3
2a2 5a = 3
2a2 5a 3 = 0
(2a + 1) (a
3) = 0
1
2
or
cos x =
The solution of cos x =
1
is x =
2
c copyright Hidegkuti, Powell, 2009
=)
a=
1
or a = 3
2
cos x = 3
2
+ 2k . There is no solution of cos x = 3.
3
Last revised: May 26, 2010
Trigonometric Equations 2
Lecture Notes
3.) 3 (1
page 4
sin x) = 2 cos2 x
sin2 x, this equation wil become quadratic in sin x:
Solution: After we write cos2 x = 1
3 (1 sin x)
3 (1 sin x)
3 3 sin x
2
2 sin x 3 sin x + 3
2 sin2 x 3 sin x + 1
(2 sin x 1) (sin x 1)
sin x =
=
=
=
=
=
=
2 cos2 x
2 1 sin2 x
2 2 sin2 x
2
0
0
1
2
or
cos2 x = 1 sin2 x
distribute
add 2 sin2 x
subtract 2
factor
sin x = 1
5
1
is x = + 2k and x =
+ 2k where k 2 Z. The solution of sin x = 1 is
2
6
6
+ 2k where k 2 Z.
The solutoin of sin x =
x=
2
4.) sin x =
cos x
Solution: Since neither sin x nor cos x is squared, it is di¢ cult to eliminate either one of them. There are
several methods to solve this equation. One method involves squaring both sides of the two equation and
then eliminating one trigonometricfunction in terms of the other. Notice that, because of the squaring, this
method creates extranneous solutions so we have to carefully check our solution. The method presented
here is a clever alternative.
Case 1. If cos x = 0
If cos x = 0; then clearly sin x =
equation sin x = cos x.
1 since sin x =
p
1
cos2 x: Then x is clearly not a solution of the
Case 2. If cos x 6= 0; then we can divide by it.
sin x =
sin x
=
cos x
tan x =
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cos x
divide by cos x 6= 0
1
1
Last revised: May 26, 2010
Trigonometric Equations 2
Lecture Notes
The solution is x =
5.) tan2 x =
4
page 5
+ k where k 2 Z.
1
3
Solution:
tan2 x =
1
tan x = p
3
1
3
tan x =
1
p
3
or
tan x =
1
p
3
1
The solution of tan x = p is x = + k where k 2 Z, and the solution of tan x =
6
3
where k 2 Z.
c copyright Hidegkuti, Powell, 2009
1
p is x =
3
6
+k
Last revised: May 26, 2010
Trigonometric Equations 2
Lecture Notes
page 6
6.) cos x sin x = cos x
Solution:
cos x sin x = cos x
cos x sin x cos x = 0
cos x (sin x 1) = 0
cos x = 0 or sin x
cos x = 0
1=0
or
subtract cos x
factor out cos x
sin x = 1
+ 2k where k 2 Z (notice this can be written simpler, as x = + k )
2
2
+ 2k where k 2 Z: As the picture already indicates, the second case does
The solution of cos x = 0 is x =
and the solution of sin x =
2
not bring in any new solutions; if sin x = 1 then of course also cos x = 0. The …nal answer is x = + k
2
where k 2 Z.
7.) tan x = tan2 x
Solution:
tan x = tan2 x
0 = tan2 x tan x
0 = tan x (tan x 1)
tan x = 0
or
subtract tan x
factor out tan x
tan x = 1
The solution of tan x = 0 is x = k where k 2 Z, and the solution of tan x = 1 is x =
c copyright Hidegkuti, Powell, 2009
4
+ k where k 2 Z.
Last revised: May 26, 2010
Trigonometric Equations 2
Lecture Notes
8.) 2 cos x
page 7
sin x + 2 cos x sin x = 1
Solution:
2 cos x sin x + 2 cos x sin x
2 cos x sin x + 2 cos x sin x 1
2 cos x (sin x + 1) (sin x + 1)
(2 cos x 1) (sin x + 1)
=
=
=
=
1
0
0
0
1 = 0 or sin x + 1 = 0
1
cos x =
or
sin x =
2
2 cos x
subtract 1
factor by grouping
1
9.) cos x = 1 + sin2 x
Solution:
cos x
cos x
2
cos x + cos x
2
cos x + cos x 2
(cos x 1) (cos x + 2)
=
=
=
=
=
cos x = 1
1 + sin2 x
1 + 1 cos2 x
2
0
0
or
sin2 x = 1 cos2 x
add cos2 x
subtract 2
factor
cos x =
2
The solution of cos x = 1 is x = 2k where k 2 Z, and the equation cos x =
c copyright Hidegkuti, Powell, 2009
2 has no solution.
Last revised: May 26, 2010
Trigonometric Equations 2
Lecture Notes
page 8
10.) 7 sin x + 5 = 2 cos2 x
Solution:
7 sin x + 5
7 sin x + 5
7 sin x + 5
2
2 sin x + 7 sin x + 5
2 sin2 x + 7 sin x + 3
(2 sin x + 1) (sin x + 3)
sin x =
The solution of sin x =
has no solution.
1
is x =
2
6
=
=
=
=
=
=
1
2
2 cos2 x
2 1 sin2 x
2 2 sin2 x
2
0
0
or
+ 2k and x =
sin x =
cos2 x = 1 sin2 x
distribute
add 2 sin2 x
subtract 2
factor
3
5
+ 2k where k 2 Z, and the equation cos x =
6
3
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c copyright Hidegkuti, Powell, 2009
Last revised: May 26, 2010