Solving a Trig Equation (Correct?)

[whyorwhynot]

Homework Statement

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

Homework Equations

cos²x + sin²x = 1

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

 

[Tedjn]

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

 

[tiny-tim]

Hi CaptainADHD!

Please don't give out complete answers on this forum.

 

[CaptainADHD]

Hi CaptainADHD!

Please don't give out complete answers on this forum.

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.

 

[tiny-tim]

Can you give me the power to edit my old posts or something?

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

 

[Дьявол]

I will help you a little bit.

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)

 

[JANm]

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

 

[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.

 

[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).