Solve cos 4x + cos x = 0, 0°<x<360°?

Question

I got stucked after getting the next step: 2cos² 2x - 1 + cos x = 0, and I do not know how to continue..

Should I change the cos² 2x to (2cos² x -1)² ? But it looks more complicated..

Answer 1

cos(A+B) = cosAcosB - sinAsinB
cos(A-B) = cosAcosB + sinAsinB

cos(A+B) + cos(A-b) = 2cosAcosB

A+B = 4x
A-B = x
2A = 5x
A = 5x/2
B = 3x/2

cos 4x + cos x
= 2cos(5x/2)cos(3x/2)

2cos(5x/2)cos(3x/2) = 0
cos(5x/2)cos(3x/2) = 0

cos(5x/2) = 0
5x/2 = {90°, 270°, 450°, 630°, 810°}
x = {36°, 108°, 180°, 252°, 324°}

cos(3x/2) = 0
3x/2 = {90°, 270°, 450°}
x = {60°, 180°, 300°}

x = {36°, 60°, 108°, 180°, 270°, 300°, 324°}

Answer 2

Use the identity
cos A + cos B = 2 cos [(A + B)/2] sin [(A - B)/2]

So cos 4x + cos x = 2 cos 5x/2 sin 3x/2
0 = 2 cos 5x/2 sin 3x/2

Case 1: When cos 5x/2 = 0
0 = cos 90 = cos 270 = cos 450 = cos 630 = cos 810 = cos 990 = ...
Now put each of the above angles equal to 5x/2 and solve for x till you get x > 360
5x/2 = 90 or x = 36
5x/2 = 270 or x = 108
5x/2 = 450 or x = 180
5x/2 = 630 or x = 252
5x/2 = 810 or x = 324
5x/2 = 990 or x = 396 which is not in accordance with the question. So we stop here.

Case 2: When sin 3x/2 = 0
0 = sin 0 = sin 180 = sin 360 = sin 540 = sin 720 = ...
Proceed as in Case 1.
3x/2 = 0 or x = 0
3x/2 = 180 or x = 120
3x/2 = 360 or x = 240
3x/2 = 540 or x = 360
3x/2 = 720 or x = 480. Here we stop.

So x = 0, 36, 108, 120, 180, 240, 252, 324, 360.

Hope this helps.

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Answer 3

sure!!!!! no, not really.

You want the difference to be 0 so one cosine has to be the negative of the other, right?

Also cosine is positive in the first and fourth quadrants but lets think about x being in the first. If x is in the first then the negative will be in the second or third and the angle will be equal to either pi - x or pi + x (draw it out, it should be clear).

4x = pi - x so 5x = pi and x = pi/5
4x = pi +x so 3x = pi and x = pi/3

What about the fourth:
cos(2pi - x) = cos(x)

So x = 9pi/5 and 5pi/3

There are also the second and third quadrants.
4x = 2pi + pi - x ... x in the second 4x in the first
5x = 3pi or x = 3pi/5

4x = 4pi - pi + x = pi + x
3x = 3pi and x = pi

Third quadrant:
4x = 8pi - pi - x and x = 7pi/5

All: pi/5, pi/3, 3pi/5, pi, 7pi/5,5pi/3,9pi/5

Answer 4

cos 4x + cos x = 0

We use the sum of cosines formula:
cosA + cosB = 2 cos [(A+B) / 2 ] cos [(A-B) / 2 ]

Then:
2 cos [(4x+x) / 2 ] cos [(4x-x) / 2 ] = 0
2 cos (5x/2) * cos (3x/2)= 0

So:
1st) cos (5x/2) = 0
or
2nd) cos (3x/2)= 0

-------------------------
1st)
cos (5x/2) = 0
5x/2 = 90º + k*180
5x = 2(90º + k*180)
x = 2(90º + k*180) / 5
x = 36 + k*72

Then:
x1 = 36
x2 = 36 + 72 = 108
x3 = 36 + 2*72 = 180
x4 = 36 + 3*72 = 252
x5 = 36 + 4*72 = 324

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2nd)
cos (3x/2) = 0
3x/2 = 90º + k*180
3x = 2(90º + k*180)
x = 2(90º + k*180) / 3
x = 60 + k*120
Then:
x6 = 60
x7 = 60 + 120 = 180
x8 = 60 + 2*120 = 300

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Solutions:
x = 36
x = 60
x = 108
x = 180
x = 252
x = 300
x = 324
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