rachelmaddie
Full Member
- Joined
- Aug 30, 2019
- Messages
- 851
-cosx - cosx = - 2Can't you just simplify the LHS using the fact that A + A = 2A? Then divide by 2, and so on. It seems trivial to me -- in fact, too trivial to be a real problem!
And the solution of that problem is not what you say. Was there a typo somewhere?
Yes - but your OP asked for solutions - plural. So what are the other solution/s?-cosx - cosx = - 2
-2cosx = -2
cosx = 1
x = 0
Like this?
the general solution is 360 × nYes - but your OP asked for solutions - plural. So what are the other solution/s?
Which of course means that the only solution in the stated interval is 0.the general solution is 360 × n
Can you please check this?Which of course means that the only solution in the stated interval is 0.
It's worth noting that in this context, the plural is hypothetical! It is possible for "all solutions" to be one solution, or even none.
By the way, I solved it a little differently: [MATH]\sin(3\pi/2 + x) = -1[/MATH], so [MATH]3\pi/2 + x = 3\pi/2 + 2n\pi[/MATH], and [MATH]x = 2n\pi[/MATH]. (You gave this in terms of degrees, which is of course wrong, as the equation is written with radians.)
Like this?You don't need to show both your solution and mine; either alone is enough!
To show that your solution, x = 0, is the only one in the given interval, you just have to know that the cosine is only 1 at one place in a cycle, namely the beginning (which is the same as the end, the start of the next cycle).
Alternatively, you could do as I did and state explicitly the general solution and then point out that only one solution is in the first cycle:
-cosx - cosx = - 2-2cosx = -2cosx = 1x = 0 + 2n pi (for any integer n) -- general solutionx = 0 -- only solution in [0, 2 pi)
No. Drop the line about the sine, which doesn't belong there at all. That is my alternate solution, not part of yours.Like this?
-cosx - cosx = - 2
-2cosx = -2
cosx = 1
sin(3pi/2 + x) = -1, so 3pi/2 + x = 3pi/2 + 2npi, and x = 2npi
x = 0 + 2n pi (for any integer n) -- general solution
x = 0 -- only solution in [0, 2 pi)
-cosx - cosx = - 2No. Drop the line about the sine, which doesn't belong there at all. That is my alternate solution, not part of yours.
Please think about how the lines you write fit together, rather than blindly copying things as given to you and sticking them together. The goal here is understanding, and you seem to be avoiding that.