# Solving a Trig Equation (Correct?)

1. Nov 3, 2008

### whyorwhynot

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

2. Nov 3, 2008

### Tedjn

Recheck your third statement. -cos2x is not equal to cos2x except when the cosine is 0.

3. Nov 4, 2008

### tiny-tim

4. Nov 5, 2008

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.

5. Nov 5, 2008

### tiny-tim

I think there's only a 24-hour "edit window".

6. Nov 7, 2008

### Дьявол

cos2x=1-sin2x

Substitute for cos2 in sinx-cos2x-1. After doing the mathematical operations, you will get quadratic equation at the end which you need to find out, and write in the form: a(x-x1)(x-x2), where a is the coefficient before y2
(ay2+by+c=0)

7. Nov 12, 2008

### JANm

Strange a quadratic equation with seemingly two solutions, but only one works?

8. Nov 13, 2008

### Дьявол

@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.

9. Nov 14, 2008

### JANm

So we get sin^2(x)+sin(x)-2,
(sin(x)-x_1)*(sin(x)-x_2),
x_1,2=(-1+/-sqrt(1+8))/2=> x1=1, x2=-2,
the factorisation is (sin(x)-1)*(sin(x)+2).