# find the number of solutions of the equation `cos^2x-1=0` for the values of x in the interval (0,2╥). a. 4 b. 3 c. 2 d. 1 e. 0

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`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).*

The graph for the equation is as follows:

Factoring: `c``os^2x-1= (cosx+1)(cosx-1)=0`

` ` Solving: `cosx=-1 `

`and cosx=1`

`x=acos(-1)=pi`

`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.**