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find the number of solutions of the equation `cos^2x-1=0` for the values of x in the...
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The graph for the equation is as follows:
Factoring: `c``os^2x-1= (cosx+1)(cosx-1)=0`
` ` Solving: `cosx=-1 `
`x=acos(1)=0 or x=2pi`
For the interval between 0 and 2`pi` not including 0 and not including 2`pi`, the only value of x included is `pi` .
Therefore the answer is d:
The number of solutions of `cos^2x-1=0` in the interval
(`0,2pi` ) is 1.
Posted by flbyrne on August 4, 2013 at 12:25 AM (Answer #1)
`cos^2x-1 = 0`
`(cosx-1)(cosx+1) = 0`
`cosx-1 = 0` OR `cosx+1 = 0`
`cosx-1 = 0`
`cosx = 1`
`cosx = cos0`
`x = 2npi+-0` where `n in Z`
n = 0 then `x = 0`
n = 1 then `x = 2pi`
`cosx+1 = 0`
`cosx = -1`
`cosx = cospi`
`x = 2mpi+-pi` Where `m in Z`
m = 0 then `x = +-pi`
m = 1 then `x = 3pi` OR `x = pi`
m = 2 then `x = 5pi` OR `x = 3pi`
In the question the answer is required at the interval `(0,2pi)` . This means `x = 0` or `x = 2pi` cannot be a answer.
So the answers are;
`x = pi`
We have only one solution. The answer is d).
Posted by jeew-m on August 4, 2013 at 12:55 AM (Answer #2)
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