Question

How do you solve `cscx+cotx=1` and find all solutions in the interval `[0,2pi)`?

Answer 1

`1/sinx + cosx/sinx =1`

`(1 + cosx)/sinx = 1/1`

`1+ cosx = sinx`

`(1 + cosx)^2 = (sinx)^2`

`1 + 2cosx + cos^2x = sin^2x`

`1 + 2cosx + cos^2x = 1 - cos^2x`

`2cos^2x + 2cosx + 1 - 1 = 0`

`2cosx(cosx + 1) = 0`

`cosx = 0 and cosx = -1`

`x= pi/2, (3pi)/2, pi`

However, checking in the original equation, you will find that `x = (3pi)/2` is extraneous and `x = pi` makes the equation undefined and is thus also extraneous. Hence, the only solution in the interval `[0, 2pi)` is `{pi/2}`.

Hopefully this helps!

Answer 2

`color(red)("solution is "x=pi/2)`

Explanation:

Given equation

`cscx+cotx=1....(1)`

Now we know

`csc^2x-cot^2x=1`

`=>(cscx+cotx)(cscx-cotx)=1`

`=>1*(cscx-cotx)=1`

`=>(cscx-cotx)=1....(2)`

Adding (1) & (2) we get

`2cscx=2=>cscx=1=csc(pi/2)`

`:.x=pi/2`

Again Subtracting (2) from (1)

`2cotx=0=>cotx=0=cot(pi/2)`
`:.x=pi/2`

For `cotx=0=cot(3pi/2)`
then `x=3pi/2` but this does not satisfy the given equation as `sin(3pi/2)=-1`

`color(red)("So only solution is "x=pi/2)`

Alternative

Given equation

`cscx+cotx=1`

`=>1/sinx+cosx/sinx=1`

`=>(1+cosx)/sinx=1`

`=>sinx-cosx=1`

`=>1/sqrt2*sinx-1/sqrt2*cosx=1/sqrt2`

`=>sin(pi/4)*sinx-sin(pi/4)*cosx=1/sqrt2`

`=>sin(x-pi/4)=sin(pi/4)`

`=>x=pi/4+pi/4=pi/2`

Again

`=>sin(x-pi/4)=1/sqrt2=sin(3pi/4)`

`=>x=(3pi)/4+pi/4=pi`
But this does not satisfy the given equation as `cscpi and cot pi" undefined"`

`color(red)("So only solution is "x=pi/2)`