Trigonometric Equation solving

[bnosam]

Either you're not reading your book correctly, or the book has a typo in a very important identity.
It should read cos 2x = cos²(x) -[/color] sin²(x)

That's "minus" NOT "equals".

Yeah, it was a typo on their part I guess. I'll redo the question.

continuing from after sinx = 02cos²x + sin²x -1 = 0

2(1-2sin²x) + sin²x - 1 = 0
1 - 3sin²x = 0
3sin²x = 1

sin²x = 1/3

I get the same answer? I must be doing something wrong.

 

[Mark44]

Yeah, it was a typo on their part I guess.

If it was my book, I would replace that '=' with '-' using a heavy pen.

I'll redo the question.

continuing from after sinx = 0


2cos²x + sin²x -1 = 0

2(1-2sin²x) + sin²x - 1 = 0

The above is wrong. You replaced cos²(x) with 1 - 2sin²(x).

The identity you should be thinking of is sin²(x) + cos²(x) = 1. If you solve for cos²(x) in this identity, what do you get?

1 - 3sin²x = 0
3sin²x = 1

sin²x = 1/3

I get the same answer? I must be doing something wrong.

 

[bnosam]

If it was my book, I would replace that '=' with '-' using a heavy pen.
The above is wrong. You replaced cos²(x) with 1 - 2sin²(x).

The identity you should be thinking of is sin²(x) + cos²(x) = 1. If you solve for cos²(x) in this identity, what do you get?

I need to pay more attention reading the identities, sorry.

sin3x = sinx
sin(2x + x) = sin2x cosx + sinx cos2x
(2sinxcosx)cosx + sinx(2cos²x - 1) = sinx
2sinx cos²x + 2cos²x sinx - 2 sinx
4sinxcos²x - 2sinx = 0

sinx(2cos²x - 1) = 0

sinx = 0


2cos²x -1 = 0
cos²x = 1/2
cos x = ± √2/2

 

[Mark44]

I need to pay more attention reading the identities, sorry.

sin3x = sinx
sin(2x + x) = sin2x cosx + sinx cos2x Put this line to the side. It just explains which identity you're using.[/color]
(2sinxcosx)cosx + sinx(2cos²x - 1) = sinx
2sinx cos²x + 2cos²x sinx - 2 sinx = 0[/color]
4sinxcos²x - 2sinx = 0

sinx(2cos²x - 1) = 0

sinx = 0

OR[/color]
2cos²x -1 = 0
cos²x = 1/2
cos x = ± √2/2

This looks good, now, but you're not done yet.

If sin(x) = 0, x = ? With no restrictions, there are an infinite number of values.
Or if cos(x) = ± √2/2, x = ? As above, with no restrictions given, there are an infinite number of values.

 

[bnosam]

This looks good, now, but you're not done yet.

If sin(x) = 0, x = ? With no restrictions, there are an infinite number of values.
Or if cos(x) = ± √2/2, x = ? As above, with no restrictions given, there are an infinite number of values.

The interval of values to solve for is 0° ≤ x ≤ 360°.

so x in sin x= 0 can be equal to 0°, 180° and 360° in that interval


x in cos x = ±√2/2

so 45°, 135°, 225°, 315° are valid

 

[Mark44]

Looks good.

A good thing to do is to check to confirm that sin(3x) = sin(x) for each of the seven values you have.

 

[epenguin]

Haven't followed this but you said trig identities. The trig identity to me most obviously relevant to problem #1 is the one for cos2x in terms of cos x, have you tried that?

(I think that gives you one of your (families of) solutions, and there is only one other real... if it's 20° as was said, that's' nothing very special or interesting to spend your time on IMHO).