What Are the Solutions for cos(2x) + cos(x) = 0?

AI Thread Summary
The equation cos(2x) + cos(x) = 0 is solved within the range of 0 to 360 degrees, yielding solutions at x = 60 degrees and x = 180 degrees, with a potential additional solution at x = 300 degrees. The discussion highlights the importance of correctly interpreting the notation, emphasizing the use of parentheses for clarity. Participants suggest rewriting cos(2x) as 2cos²(x) - 1 to facilitate solving a quadratic equation. There is a consensus that all possible values for x have been identified, although initial confusion about notation was addressed. Overall, the conversation underscores the significance of clear mathematical expression in problem-solving.
Helly123
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Homework Statement


cos2x + cos x = 0 (0 <= x <= 360)

Homework Equations

The Attempt at a Solution


cos2x + cos x = 0
2cos(3x)/2 cos(x)/2 = 0

3x/2 = 90 degrees
x = 60 degrees

x/2 = 90
x = 180

3x/2 = 270
x = 180

x/2 = 270
x = 540 (not qualified)

is there any more possibility (answers) for x?
 
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Helly123 said:
cos2x + cos x = 0 (0 <= x <= 360)
Is it cos(2x) or cos2x?
 
cnh1995 said:
Is it cos(2x) or cos2x?
it's cos2x
 
Helly123 said:
it's cos2x
Of course:-p!
I hadn't read your complete solution.

But I think you've found all the possible values for x.
 
cnh1995 said:
Of course:-p!
I hadn't read your complete solution.

But I think you've found all the possible values for x.
thank you. But there's one more answers but I don't know how to find it
 
Check the 270...
 
BvU said:
Check the 270...
I see, I get it. thanks
 
Helly123 said:

Homework Statement


cos2x + cos x = 0 (0 <= x <= 360)

Homework Equations

The Attempt at a Solution


cos2x + cos x = 0
2cos(3x)/2 cos(x)/2 = 0

3x/2 = 90 degrees
x = 60 degrees

x/2 = 90
x = 180

3x/2 = 270
x = 180

x/2 = 270
x = 540 (not qualified)

is there any more possibility (answers) for x?
What about 300°?
 
Why didn't you just write $$\cos{2x}=2\cos^2{x}-1$$and then directly solve the resulting (factorable) quadratic equation for ##\cos{x}##
 
  • #10
ehild said:
What about 300°?
Yes, that's the answer
 
  • #11
Chestermiller said:
Why didn't you just write $$\cos{2x}=2\cos^2{x}-1$$and then directly solve the resulting (factorable) quadratic equation for ##\cos{x}##
I see.. I didn't realize it
 
  • #12
Helly123 said:
cos2x + cos x = 0 (0 <= x <= 360)

cnh1995 said:
Is it cos(2x) or cos2x?
To prevent such confusion, use parentheses -- i.e., write cos(2x) instead of cos2x. Also, we get a lot of students who are new to writing math expressions as text. I've seen many who write cos2x when they really mean cos2(x). For exponents, the ^ symbol is often used.
 
  • #13
Mark44 said:
To prevent such confusion, use parentheses -- i.e., write cos(2x) instead of cos2x. Also, we get a lot of students who are new to writing math expressions as text. I've seen many who write cos2x when they really mean cos2(x). For exponents, the ^ symbol is often used.
Ok I'll be more careful
 
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