Question

(PLEASE EXPLAIN) How to Solve Tanx Cosx Sinx-1=0 algebraically, over 0 ≤ x <360 Deg.?

The answer is: No solution

But I want to know why is that the answer.

Answer 1

See explanation.

Explanation:

`tan(x)=sin(x)/cos(x)`, so

`tan(x)cos(x)sin(x)-1=0` can become

`sin(x)/cos(x)\*cos(x)\*sin(x) = 1`

which can simplify to

`sin^2(x) = 1`

(but we canceled out a denominator here, which is key!)

`sin^2(x)=1\rightarrow sin(x)=1` or `sin(x)=-1`

`sin(x)=1` gives `x=90^circ`

`sin(x)=-1` gives `x=270^circ`

The issue with both of these answers is that `tan(x)` does not exist at those values and we always go back to the original to check our solutions. Since they don't work in the original they are extraneous solutions and there are no solutions to the given equation.