Разложите на множители 4x^2y^2-9 и 125-0, 064p^3 ​

1)2-2cos²x+11cosx-7=0 2cos²x-11cosx+5=0 cosx=a 2a²-11a+5=0 d=121-40=81 a1=(11-9)/4=1/2⇒cosx=1/2⇒x=+-π/3+2πn a2=(11+9)/4=5⇒cosx=5 нет решения 2)tgx=a 3a²-7a+2=0 d=49-24=25 a1=(7-5)/6=1/3⇒tgx=1/2⇒x=arctg1/3+πn a2=(7+5)/6=2⇒tgx=2⇒x=arctg2+πn 3) 4cosx+3sinx=04cos²x/2-4sin²x/2+6sinx/2cosx/2=0/cos²x/2≠02tg²x/2-3tgx/2-2=0tgx/2=a2a²-3a-2=0d=9+16=25a1=(3-5)/4=-1/2⇒tgx/2=-1/2⇒x/2=-arctg1/2+πn⇒x=-2arctg1/2+πna2=(3+5)/4=2 ⇒tgx/2=2⇒x/2=arctg12+πn⇒x=2arctg2+πn 4)3sin^2x-5sinx=0sinx(3sinx-5)=0sinx=0⇒x=πnsinx=5/3 нет решения 5)sin2x+sinx=0 2sinxcosx+sinx=0 sinx(2cosx+1)=0 sinx=0⇒x=πncosx=-1/2⇒x=+-2π/3+2πn

324 - ( z - 11)^2 = 324 - ( z^2 - 22z + 121 ) = 324 - z^2 + 22z - 121 = = - z^2 + 22z+ 203 = z^2 - 22z - 203 = ( z - 29)*( z + 7 ) d = 484 + 812 = 1296 ;   √ d = 36  z1 = ( 22 + 36 ) : 2 = 29  z2 = ( 22 - 36 ) : 2 = - 7  ответ ( z - 29)*( z + 7 ) ================================================== 256 - ( b + 1)^2 = 256 - b^2 - 2b - 1 = - b^2 - 2b + 255 = b^2 + 2b - 255 = = ( b - 15)*( b + 17 ) d = 4 + 1020 = 1024 ;   √ d = 32  b1 = ( - 2 + 32 ) : 2 = 15  b2 = ( - 2 - 32 ) : 2 = - 17  ответ ( b - 15)*( b + 17 )