1)sin^2x-4sinxcosx+3cos^2x=0
2)sin5xsin4x+cos6xcos3x=0
3)2sin^2x+3sinxcosx-2cos^2x=0
4)cos6xcos12x=cos8xcos10x

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вопрос

1) Given equation can be rewritten as:
(sin^2x + 3cos^2x) - 4sinxcosx = 0
3 - 4sinxcosx = 0
3 = 4sinxcosx
Divide both sides by 4:
3/4 = sinxcosx
sin2x = 3/4

2) Using trigonometric identity sin(a)sin(b) = (1/2)cos(a-b) - (1/2)cos(a+b) and cos(a)cos(b) = (1/2)cos(a-b) + (1/2)cos(a+b), we get:
sin(5x-4x) - sin(5x+4x) + cos(6x-3x) + cos(6x+3x) = 0
sinx - sin9x + cos3x + cos9x = 0

3) Given equation is a quadratic equation:
2sin^2x + 3sinxcosx - 2cos^2x = 0
(2sinx - cosx)(sinx + 2cosx) = 0
sinx = 1/2, -2

4) Using trigonometric identity cos(a)cos(b) = (1/2)[cos(a-b) + cos(a+b)], the equation can be rewritten as:
(1/2)(cos6x-12x + cos6x+12x) = cos8xcos10x
(1/2)(cos6xcos12x) = cos8xcos10x
cos6x = cos8xcos10x
This could be true only when sin6x=0.
So, the equation cos6xcos12x=cos8xcos10x holds true when sin6x=0.