Amath HELP! (angle)

1
In triangle ABC, cos2A + cos2B +cos2C = -1-4cosAcosBcosC
SOLUTION
A+B+C=π，that is C=π-(A+B)．

∴
LHS
=2cos(A+B)cos(A-B)+2 cos^2C-1

=2cos(A+B)cos(A-B)+2 cos^2(A+B)-1

=2cos(A+B)[cos(A+B)+cos(A-B)]-1
=4cos(A+B)cosAcosC-1
=-1-4cosAcosBcosC
=RHS
2.) (a)

Using the identity 2sinAcosB = sin(A+B) + sin(A-B),

prove that 2 (sinx - sin3x + sin5x - sin7x + sin9x) cosx = sin10x

(b)Hence or otherwise, solve sinx + sin5x + sin9x = sin3x + sin7x
(a)
LHS
=2 (sinx - sin3x + sin5x - sin7x + sin9x) cosx
=2sinxcosx-2sin3xcosx + 2sin5xcosx - 2sin7xcosx + 2sin9xcosx
=(sin2x)-(sin4x+sin2x)+(sin6x+sin4x)-(sin8x+sin6x)+(sin10x+sin8x)
=sin10x
(b)
sinx + sin5x + sin9x = sin3x + sin7x
sinx - sin3x + sin5x - sin7x + sin9x = 0
2 (sinx - sin3x + sin5x - sin7x + sin9x) cosx =0
sin10x=0 (by the result of (a))
10x=180n (where n is an integer)
x=18n (where n is an integer)