Step by step solution

[QwertyKnight]

I need to know the step by step solution of this equation: 1+sinx-sin2x=cosx+cos2x-cos3x, because i only could found 1 solution of 5.
I'd appreciate the help!

 

[ksdhart]

Well, you say there's five solutions, but you only found one. That means you've already done a good amount of work on this problem. If you post the work you've done so far, it will help us know how to help you find the other answers. Also, unless the problem specified that you're only looking at x over some interval (typically 0 to 360 degrees), then the problem, in fact, has infinite solutions, of five general forms.

 

[QwertyKnight]

Okay, i forgot to say there's 5 kind of solutions.
Here is my work:
1+sinx-sin2x=cosx+cos2x-cos3x
move cosx and cos3x to the left side
1+sinx-sin2x+cos3x-cosx=cos2x
cos3x-cosx=-2sin2x*sinx and cos2x=cos^2x-sin^2x and 1=sin^2x+cos^2x
subtract 1 from both sides
sinx-sin2x-2sin2x*sinx=cos^2x-sin^2x-(sin^2x+cos^2x)=>cos^2x-sin^2x-sin^2x-cos^2x
sinx-sin2x-2sin2x*sinx=-2sin^2x
add -2sin^2x for both sides
2sin^2x+sinx-sin2x-2sin2x*sinx=0
sin2x=2sinxcosx
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sinx*(2sinx+1-2cosx-2sin2x)=0 if sinx=0 x=k*pi
or 2(sinx-cosx-sin2x)+1=0

 

[stapel]

Okay, i forgot to say there's 5 kind of solutions.
Here is my work:
1+sinx-sin2x=cosx+cos2x-cos3x
move cosx and cos3x to the left side
1+sinx-sin2x+cos3x-cosx=cos2x
cos3x-cosx=-2sin2x*sinx and cos2x=cos^2x-sin^2x and 1=sin^2x+cos^2x...

I think you're doing something like the following:

The left-hand side contains the difference "cos(3x) - cos(x)". Applying the Cosine Difference Identity, we get:

. . .cos(3x) - cos(x) = (-2) * sin[(3x + x)/2] * sin[(3x - x)/2] = -2 sin(4x/2) sin(2x/2) = -2 sin(2x) sin(x)

The right-hand side contains the term "cos(2x)", to which we may apply the Double-Angle Identity for cosines. This gives us:

. . .cos(2x) = cos^2(x) - sin^2(x)

The left-hand side also contains the constant "1", which may be restated by using the basic Pythagorean Identity:

. . .1 = sin^2(x) + cos^2(x)

Doing the various substitutions on the equation gives us:

. . .1 + sin(x) - sin(2x) + [cos(3x) - cos(x)] = cos(2x)

. . .{ sin^2(x) + cos^2(x) } + sin(x) - sin(2x) + { -2 sin(2x) sin(x) } = { cos^2(x) - sin^2(x)}

. . .sin(x) - sin(2x) + sin^2(x) + cos^2(x) - 2 sin(2x) sin(x) = cos^2(x) - sin^2(x)

We can subtract "cos^2(x)" from, and add a "sin^2(x)" to, either side of the equation:

. . .sin(x) - sin(2x) + 2 sin^2(x) - 2 sin(2x) sin(x) = 0

We can apply the Double-Angle identity for sines twice, which gives us:

. . .sin(x) - 2 sin(x) cos(x) + 2 sin^2(x) - 2*2 sin(x)cos(x) sin(x) = 0

. . .sin(x) - 2 sin(x) cos(x) + 2 sin^2(x) - 4 sin^2(x) cos(x) = 0

However, I'm not clear on what you're doing after that...? You seem to have factored out a sine, leaving you with:

. . . . .sin(x) [1 - 2 cos(x) + 2 sin(x) - 4 sin(x) cos(x)] = 0

This means that either the sine is zero, or else:

. . . . .1 - 2 cos(x) + 2 sin(x) - 4 sin(x) cos(x) = 0

If the sine is zero, then the value of x is the product of pi and an integer. Otherwise:

Re-group terms on the left-hand side as:

. . . . .1 + 2 sin(x) - 2 cos(x) - 4 sin(x) cos(x) = 0

Group the last two terms together, and factor out a -2cos(x), leaving:

. . . . .1 + 2 sin(x) - 2 cos(x) [1 + 2 sin(x)] = 0

Note that "1 + 2 sin(x)" is now a common term, so we can do as follows:

. . . . .1 [1 + 2 sin(x)] - 2 cos(x) [1 + 2 sin(x)] = 0

. . . . .[1 + 2 sin(x)] [1 - 2 cos(x)] = 0

This gives us two equations to solve:

. . . . .1 + 2 sin(x) = 0, or sin(x) = -1/2

. . . . .1 - 2 cos(x) = 0, or 1/2 = cos(x)

Solve these.

 

[QwertyKnight]

Thx the help! I realized that, I'm not really good at factoring.