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Hey,

So... I have this problem:

-sin(x)-2cos(2x)

And I need to solve for all the x values between 0 and 2pi.

Naturally, one would equate that to zero and get:

-sin(x)-2cos(2x)=0

And in order to find what that's equivalent to, you'd have to find a way to factor it... And in order to do that, you'd have to use identities... I know the following for cos(2x):

[tex]\cos(2x)=\cos(x)^2-\sin(x)^2[/tex]

[tex]\cos(2x)=2\cos(x)^2-1[/tex]

[tex]\cos(2x)=1-2\sin(x)^2[/tex]

However, I don't see any of those actually helping. By using the last one, I get:

[tex]4\sin(x)^2-\sin(x)-2[/tex]

But there's no reasonable way to factor that. I went through the depths of complete the square hell to find this:

[tex]\frac{1}{2}(2\sin(x)-1)(4\sin(x)-1)-\frac{3}{2}=0[/tex]

[tex](2\sin(x)-1)(4\sin(x)-1)=3[/tex]

And if I set each parenthesis equal to 3, which I'm pretty sure you can't do, I get:

[tex]\arcsin(2)[/tex] and [tex]\arcsin(1/2)[/tex]

Both of which are wrong...

Can anyone shed any light on this?

Haha, I'm having quite a bit of difficulty

So... I have this problem:

-sin(x)-2cos(2x)

And I need to solve for all the x values between 0 and 2pi.

Naturally, one would equate that to zero and get:

-sin(x)-2cos(2x)=0

And in order to find what that's equivalent to, you'd have to find a way to factor it... And in order to do that, you'd have to use identities... I know the following for cos(2x):

[tex]\cos(2x)=\cos(x)^2-\sin(x)^2[/tex]

[tex]\cos(2x)=2\cos(x)^2-1[/tex]

[tex]\cos(2x)=1-2\sin(x)^2[/tex]

However, I don't see any of those actually helping. By using the last one, I get:

[tex]4\sin(x)^2-\sin(x)-2[/tex]

But there's no reasonable way to factor that. I went through the depths of complete the square hell to find this:

[tex]\frac{1}{2}(2\sin(x)-1)(4\sin(x)-1)-\frac{3}{2}=0[/tex]

[tex](2\sin(x)-1)(4\sin(x)-1)=3[/tex]

And if I set each parenthesis equal to 3, which I'm pretty sure you can't do, I get:

[tex]\arcsin(2)[/tex] and [tex]\arcsin(1/2)[/tex]

Both of which are wrong...

Can anyone shed any light on this?

Haha, I'm having quite a bit of difficulty

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