Решить уравнение: 4sin3x+1/3cos3x=3

4sin3x+1=9co3x 8sin3x/2 * cos3x/2 +sin²3x/2 + cos²3x/2 -9cos²3x/2 + 9sin²3x/2=0 10sin²3x/2 + 8sin3x/2 * cos3x/2 -8cos²3x/2=0 / cos²3x/2≠0 10tg²3x/2+8tg3x/2-8=0 tg3x/2=a 10a²+8a-8=0 d=64+320=384    √d=8√6 a1=(-8-8√6)/20=(-2-2√6)/5⇒tg3x/2=(-2-2√6)/5⇒3x/2=-arctg(2+2√5)/5+πn⇒ ⇒x=-2/3arctg(2+2√5)/5+2πn/3 a2=(-8+8√6)/20=(-2+2√6)/5⇒tg3x/2=(-2+2√6)/5⇒3x/2=arctg(-2+2√5)/5+πn⇒ ⇒x=2/3arctg(-2+2√5)/5+2πn/3

3(4x*x+4x+1)+20x+10+3=0

12x*x+12x+3+20x+10+3=0

12x*x+32x+16=0

3x*x+8x+4=0

d=64-4*3*4=64-48=16

x1=(-8+4)/6=-4/6=-2/3

x2=(-8-4)/6=-12/6=-2

ответ: -2, -2/3

x1^2+x2^2=36

(x1+x2)-2x1x2=36

теорема виета: x1+x2=6

36-2x1x2=36

x1x2=0

x1x2=c/a=p=> p=0