ответ:
log3 = 2*log9 - 1
log3 = 2 * log(3^2) - log3 3
log3 = 2 * 1\2 * log3 - log3 3
log3 = log3 - log3 3
log3 (sin 3x - sin x) = log3 [(17*sin 2x) \ 3]
теперь основания логарифмов одинаковые =>
решать выражения при логарифмах (приравнять их):
sin 3x - sin x) = [(17*sin 2x) \ 3]
3*(sin 3x - sin x) = 17*sin 2x
3*[(3sin x - 4sin^3 x) - sin x] = 17*(2sin x * cos x)
3*(2sin x - 4sin^3 x) = 34*sin x * cos x > (: ) на sin x =>
6 - 12sin^2 x = 34cos x
6 - 12*(1 - cos^2 x) = 34cos x
6 - 12 + 12cos^2 x - 34cos x = 0
12cos^2 x - 34cos x - 6 = 0 > (: ) на 2 и cos x = t
6t^2 - 17t - 3 = 0
дальше легко
объяснение:
1) ac2-ad+c3-cd-bc2+bd= = (ac2 – ad) + (c3 –
bc2) + (bd – cd) = a·(c2 – d) + c2·(c – b) + d·(b – c) = a·(c2 – d) +
c2·(c – b) – d·(c – b) = a·(c2 – d) + c2·(c – b) – d·(c – b) = a·(c2 –
d) + (c – b)·(c2 – d) = (c2 – d)·(a + c – b)
2) mx2+my2-nx2-ny2+n-m= x2 ( m - n ) + y2 ( m - n ) - ( m - n ) = ( m-n ) (x2 + y2 - 1 )
3) am2+cm2-an+an2-cn+cn2= m2 (a + c ) + n2 ( a + c ) - n ( a + c ) = ( a+ c) ( m2 + n2 - n)
4) xy2-ny2-mx+mn+m2x-m2n= y2 ( x - n ) + m2 ( x - n) - m ( x - n ) = ( x-n) ( y2 + m2 - m )
5) a2b+a+ab2+b+2ab+2=ab ( a + b + 2 ) + ( a+ b+ 2 ) = 2 ( a+ b + 2 )
6) x2-xy+x-xy2+y3-y2= x ( x – y + 1) – y 2 ( x – y + 1)=( x – y + 1)( x – y 2 ).
