Question

How do you solve `(sinx+cosx)/tanx+(1-sinx)/sinx=cosx` for `0<=x<=2pi`?

Answer 1

`{}`; no solution

Explanation:

Rewrite `tanx` as `sinx/cosx`:

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

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

`(cosxsinx + cos^2x + 1 - sinx)/sinx = cosx`

`cosxsinx + cos^2x + 1 - sinx = sinxcosx`

`cos^2x + 1 - sinx = sinxcosx- sinxcosx`

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

Use `sin^2x + cos^2x = 1`:

`1 - sin^2x + 1 - sinx = 0`

`0 =sin^2x + sinx - 2`

Factor:

`0 = (sinx + 2)(sinx - 1)`

`sinx = -2 and 1`

There is no solution to `sinx = -2`. However, `sinx = 1` is solvable.

`x = pi/2`

However, this is extraneous, since `tan(pi/2)` is not defined in the real number system, aka:

`tan(pi/2) = sin(pi/2)/cos(pi/2) = 1/0 = "undefined"`

Therefore, this equation has no solution. This can be symbolized with an empty solution set `{}`.

Hopefully this helps!