Can you help me solve cos2x + cosx = 0 ?

Answer 1

cos2x=2cos^2(x) -1
2cos^2(x) -1 +cos(x)=0
x= 180°

Answer 2

Okay I admit I have no idea what this is... but I found this online....

Math 150 T16-Trigonometric Equations Page 1
MATH 150 { TOPIC 16
TRIGONOMETRIC EQUATIONS
In calculus, you will often have to nd the zeros or x-intercepts of a function.
That is, you will have to determine the values of x that satisfy the equation
f(x) = 0. This section reviews some techniques for solving f(x) = 0 when
f(x) is a trigonometric function. A good grasp of Review Topics 12, 13, 14,
and 15 will prove to be very useful. Also, remember that after you solve an
equation you can always check your answer.
We rst consider the following equation.
Solve for x: sinx =
1
2
, 0 x  2.
The graph of sin x for 0  x  2 and the line y =
1
2
appear below.
f(x) = sinx
−1
1 

1
2
y = 1
2

−1 p2



2
Fig. 16.1.
When 0  x  2, we see that there are two values of x where sin x =
1
2
;
namely, x =

6
and x =
5
6
. Similarly, the equation sin x = −
1
p2
has
solutions x =
5
4
and x =
7
4
. The equation sin x = 2 has no solutions
since the range of sin x = [−1; 1] (i.e., the outputs must be between −1 and
1).
Math 150 T16-Trigonometric Equations Page 2
A slightly more dicult equation to solve is the following.
Solve for x: sinx = :4, 0  x  2.
In the previous equations we were able to solve for x exactly. Now we must
use a calculator, which yields that x = arcsin(:4) = :41 radians. The angle
.41 radians is in the rst quadrant. Fig. 16.1 implies there is a second value
of x where sin x = :4. Using the concept of the reference angle (Review
Topic 9a), this second value lies in the second quadrant and = ( − :41) or
2.73 radians.
Practice Problem
16.1. Solve the following equations where 0  x  2.
a) cos x =
1
2
b) sin x = 0
c) cos x = :65 (use a calculator)
Next, consider a problem of the form:
sin 2x =
1
2
; 0  x  2:
Although we plug in x, the sine \sees" 2x. In other words, if 0  x  2,
the sine \sees" 0  2x  4. On this extended interval [0; 4], sin( ) =
1
2
at

6
,
5
6
,

6
+ 2,
5
6
+ 2. This means 2x =

6
;
5
6
;

6
+ 2;
5
6
+ 2.
Thus,
x =

12
;
5
12
;

12
+  or
13
12
;
5
12
+  or
17
12
:
Practice Problem
16.2. Solve the following equation.
cos 2x = 1; 0  x  2:
Math 150 T16-Trigonometric Equations Page 3
Many times trigonometric equations will be quadratic in form. Consider the
examples below.
2 cos2 x = 3 cos x −1 versus 2r2 = 3r − 1
sin2 x − sin x = 0 versus r2 − r = 0
This suggests solving the trigonometric equation as if it were a quadratic
equation. Let's do the rst example.
Solve: 2 cos2 x = 3 cos x − 1, 0  x  2x.
2 cos2 x − 3 cos x + 1 = 0 Get 0 on one side
(2 cos x − 1)(cos x − 1) = 0 Factor
2 cos x − 1 = 0 cos x − 1 = 0 Set each factor = 0
2 cos x = 1 cos x = 1
cos x =
1
2
x = 0; 2
x =

3
;
5
3
Solutions: x = 0,

3
,
5
3
, 2.
Practice Problem
16.3. Solve the following equations.
a) sin2 x −
3
4
= 0; 0  x  2
b) tan2 x + tan x = 0; −

2
< x <

2
.
Trigonometric identities can be involved in solving trigonometric equations.
Consider the following problem.
Solve: cos 2x + cos x = 0; 0  x  2.
Math 150 T16-Trigonometric Equations Page 4
The fact that one function has argument 2x and the other function has
argument x suggests that we try to get the same argument in both functions.
Using the identity cos 2x = 2 cos2 x − 1, we proceed as follows.
cos 2x + cos x = 0; 0  x  2
2 cos2 x + cos x − 1 = 0
(2 cos x − 1)(cos x + 1) = 0
2 cos x − 1 = 0 cos x + 1 = 0
cos x =
1
2
cos x = −1
x =

3
;
5
3
x = 
Solutions: x =

3
, ,
5
3
.
Question: Why did we choose the identity cos 2x = 2 cos2 x − 1 in the
preceding problem? There are other ways to write cos 2x. Why wouldn't
they work?
As another example, consider
sin x cos x = −
1
2
; 0  x  2:
Since sin 2x = 2sinx cos x, we have
sin 2x
2
= −
1
2
; or
sin 2x = −1; 0  x  2:
Notice that the sine \sees" 2x as in an earlier example. This means that
when 0  x  2, the sine \sees" 0  2x  4. Thus sin(2x) = −1 when
2x =
3
2
or 2x =
7
2
, which implies x =
3
4
or
7
4
.
Math 150 T16-Trigonometric Equations Page 5
Practice Problem
16.4. Solve the following equations.
a) 1 + sin x = 2 cos2 x; 0  x  2.
b) tan x = 2sinx; −

2
< x <

2
.
The next example is a bit tricky.
Solve: cos x + sinx = 0, 0  x  2.
In previous problems, it was helpful to get 0 on one side and then work with
the other side. For this equation, we instead move sin x to the right hand
side and obtain
cos x = −sin x; 0  x  2:
Now we remember that cos x and sin x are coordinates of a point P as it
moves around the unit circle (Review Topic 11). So, the equation cos x =
−sin x is equivalent to asking when the coordinates of P have the same
absolute value but dier in sign. A moment's thought reveals that x must
be
3
4
or
7
4
.
Another solution method is as follows.
Solve: cos x = −sin x, 0 x  2.
Divide each side by (−cos x) and obtain
−1 = tan x; 0  x  2; x 6=

2
;
3
2
:
NOTE: We must exclude

2
and
3
2
because tan x is not dened for these
values. Also,

2
or
3
2
do not solve the original equation. Thus we have
tan x = −1; 0  x  2; x 6=

2
;
3
2
:
Math 150 T16-Trigonometric Equations Page 6
Solutions: x =
3
4
;
7
4
.
Practice Problem
16.5. Solve the equation sin x − cos x = 0, 0  x  2.
In the previous problems, the equations were solved on some specied in-
terval. However, the interval can be unrestricted. For example,
Solve: sin x =
1
2
, for any x.
For problems like this, we must remember that the sine function is periodic
with period 2. So we rst solve the equation on the interval 0  x < 2,
and then we add multiples of 2. The solution is:
x =

6
+ 2k or
5
6
+ 2k; for any integer k
(k can be positive or negative).
Practice Problem
16.6. Solve the following equations where x is arbitrary.
a) cos2 x −
cos x
2
= 0.
b) 3 tan2 x − 1 = 0.
Math 150 T16-Trigonometric Equations Page 7
ANSWERS to PRACTICE PROBLEMS (Topic 16{Trigonometric Equations)
16.1. a) x =

3
;
5
3
b) x = 0; ; 2
c) Solve cos x = :65, 0  x  2.
Using arccos( ) on your calculator (in radian mode), you get x = :863
radians. However, cos x is also positive in the fourth quadrant where
3
2  x  2. The reference angle concept (Review Topic 9a) implies
that x = 2 − :863 = 5:42 radians.
Solutions: x = :863 or 5.42
16.2. Solve cos(2x) = 1, 0  x  2.
The cosine \sees" (2x), which means 0  (2x)  4. On this interval,
cos( ) = 1 when ( ) = 0; 2; 4. Thus, (2x) = 0; 2, or 4, which implies
x = 0; , or 2.
16.3. a) Method 1. Factor the left hand side and obtain

sin x −
p3
2
!
sin x +
p3
2
!
= 0; 0  x  2:
sin x −
p3
2
= 0 sin x +
p3
2
= 0
sin x =
p3
2
sin x = −
p3
2
x =

3
;
2
3
x =
4
3
;
5
3
Solutions: x =

3
;
2
3
;
4
3
;
5
3
Math 150 T16-Trigonometric Equations Page 8
Method 2. Solve for sin2 x.
sin2 x =
3
4
sin x = 
p3
2
: So,
sin x =
p3
2

sin x = −
p3
2
Now, continue as above.
b) Solve tan2 x + tan x = 0, −

2
< x <

2
:
tan x(tan x + 1) = 0
tan x = 0 tan x + 1 = 0
x = 0 tan x = −1
x = −

4
Solutions: x = −

4
; 0.
16.4. a) Solve
1 + sinx = 2 cos2 x; 0  x  2
= 2(1 − sin2 x)
= 2− 2 sin2 x
2 sin2 x + sinx − 1 = 0
(2 sin x − 1)(sin x + 1) = 0
2 sinx − 1 = 0 sin x + 1 = 0
sin x =
1
2
sin x = −1
x =

6
;
5
6
x =
3
2
Solutions: x =

6
;
5
6
;
3
2
.
Math 150 T16-Trigonometric Equations Page 9
b) Solve tan x = 2sinx, −

2
< x <

2
:
sin x
cos x − 2 sinx = 0
sin x

1
cos x − 2

= 0
sin x = 0
1
cos x − 2 = 0
x = 0
1
cos x
= 2
cos x =
1
2
x = −

3
;

3
Solutions: x = −

3
; 0;

3
.
16.5. Solve sin x − cos x = 0, 0  x  2.
Method 1: sin x = cosx; 0  x  2.
Since cos x and sin x are the x and y coordinates of a point P on the
unit circle, this question is equivalent to asking when the coordinates
have the same value. The answer is when P is in the rst and third
quadrants where x =

4
and
5
4
, respectively.
Method 2: Divide by cos x and obtain
tan x − 1 = 0; 0  x  2; x 6=

2
;
3
2
tan x = 1
x =

4
;
5
4
:
Math 150 T16-Trigonometric Equations Page 10
16.6. a) Solve cos2 x −
cos x
2
= 0, where x is arbitrary.
cos x

cos x −
1
2

= 0
cos x = 0 cos x −
1
2
= 0
x =

2
+ 2k, or cos x =
1
2
x =
3
2
+ 2k x =

3
+ 2k;
5
3
+ 2k
Equivalently,
x =

2
+ k
Solutions: x =

2
+ k,

3
+ 2k,
5
3
+ 2k, for any integer k.
b) Solve 3 tan2 x − 1 = 0.
tan2 x =
1
3
tan x = 
1
p3
tan x =
1
p3
tan x = −
1
p3
x =

6
+ k x = −

6
+ k
Solution: x = 

6
+k, for any integer k. (Remember that tan x
has period . This means that we only need to add multiples of
, not 2.)
Beginning of Topic Skills Assessment

Answer 3

First use the identity:

cos2x = 2cos^2(x) - 1

Your equation becomes:

2cos^2(x) + cosx - 1 = 0,

which is a quadratic equation in the variable cosx.

Using the quadratic formula with a=2,b=1,c=-1:

cos x = -1 +/- sqrt(1^2 - 4(2)(-1)) / 2(2)

cos x = 0.5 or -1

Assuming that x is between 0 and 2pi rads or 360 degrees,

x = 60 or 180 or 300 degrees

Answer 4

consider the trigo identity cos2x =2 cos^2 x– 1

Problem:

cos2x + cosx = 0

2 cos^2 x– 1 + cos x = 0

2 cos^2 x + cos x - 1 = 0

( 2 cos x - 1) ( cos x + 1) = 0

set each factor to zero:

a.
2cos x - 1 = 0
2 cos x = 1
cos x = 1/2

x = 60 degrees

b.
cos x + 1 = 0
cos x = -1

x = 180 degrees

Answer 5

cos 2x + cosx = 0

2 cos^2 x - 1 + cosx = 0

2cos^2x + 2cosx - cosx - 1 = 0

2cosx(cosx + 1) - 1(cosx + 1) = 0

(2cosx - 1)(cosx + 1) = 0

2cosx - 1 = 0 => cosx = 1/2 => pi/3 , 7pi/3,

cosx = -1 => x = pi, 3 pi

Answer 6

Are you familiar with the unit circle and the values of cosine at the "famous" angles, like 0, pi/3, pi/6, pi/2, etc.

Make one (yes make a unit cirlce if you havent already). Then look at it. Remember that cosine is the x-coordinate. Try to find an angle where the x-coordinate of that angle is positive, but the x-coord of two times that angle is equally negative. It is not hard. Actually there are two solutions. Good luck.

Answer 7

use the double angle formula and factorise cos2x = 2cos^2 x - 1 cos x (2 cos^2 x - 1) = 0 so cos x = 0 and cos x = + - 1/sqrt 2

Answer 8

2 cos ² x + cos x - 1 = 0
(2 cos x - 1)(cos x + 1) = 0
cos x = 1 / 2 , cos x = - 1
x = 60° , 300° , 180°
Sorry that this solution is so brief!

Answer 9

2(cosx)^2 + cosx - 1 =0
(2cosx - 1)(cosx + 1) =0

2cosx - 1 =0
x = 60deg, 300deg.

cosx + 1 =0
x = 180deg.

Answer 10

Buy a TI-89 calculator. It can solve it for you (costs $130).