## Solving a Trig Equation (Correct?)

1. The problem statement, all variables and given/known data
Write this expression in factored for as an algebraic expression of a single trig function (e.g., (2 sin x+3)(sin x-1):

sin x - cos2x - 1

2. Relevant equations
cos2x + sin2x = 1

3. The attempt at a solution
1) cos2x + sin2x = 1
2) sin2x = 1-cos2x
3) -cos2x = cos2x so sin2x = -cos2x+1

But the problem calls for -cos2x-1. Would the resulting function be sin2x - (-sin2x)?

4) sin x (sin x + sin x)

I'm not certain that I did it correctly
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 Recheck your third statement. -cos2x is not equal to cos2x except when the cosine is 0.

## Solving a Trig Equation (Correct?)

Well, having the rank of captain the the attention deficit and impulse control army doesn't go well with discretion.

Can you give me the power to edit my old posts or something? I don't want to get myself banned.

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 Blog Entries: 1 @JANm Look at the first post. The question is how to "factor" the whole expression, not to solve it. If the question was to solve it, than one of the solutions will worked out? Why? Because $$-1 \leq sin(x) \leq 1$$, so sin(x)=-2 will not be the solution.