(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.

1 Answer
turksvids
Jan 2, 2018

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.

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