Question

1) Решите уравнение (9-3^0, 5x-7)log2(5-2x)=0
2) Упростите уравнение -cos( π+3x)sin x + sin 3x sin (3π/2-x)

Answer 1

$(9-3^{0,5x-7)}*log_{2}(5-2x)=0$

ОДЗ : 5 - 2x > 0

- 2x > - 5

x < 2,5

$1)9-3^{0,5x-7}=0\\\\9=3^{0,5x-7}\\\\3^{0,5x-7}=3^{2}\\\\0,5x-7=2\\\\0,5x=9\\\\x=18\\\\2)log_{2}(5-2x)=0\\\\5-2x=1\\\\2x=4\\\\x=2$

Ответ : 2 ; 18

$-Cos(\pi +3x)Sinx+Sin3xSin(\frac{3\pi }{2}-x)= Cos3xSinx-Sin3xCosx=Sin(x-3x)=Sin(-2x)=- Sin2x$