Question

ГЛАВНЫЙ МОЗГ,СПАСИ,10 КЛАСС,АЛГЕБРА​

Answer 1

Я, конечно, не главный мозг, но все же.

$sin(3x-\frac{\pi }{4})\sqrt{8x-x^2-7} =0$

$8x-x^2-7\geq 0$

$-x^2+8x-7\geq 0$

Корни уравнения:

$-x^2+8x-7=0$

D $= b^2-4ac = 64 - 28 = 36$

$x_{1} = \frac{-8+6}{-2}= 1$

$x_{2} = \frac{-8-6}{-2} = 7$

То есть x∈{1;7}

$sin(3x-\frac{\pi }{4})=0$

$3x-\frac{\pi }{4} =\pi k$, k∈Z

$3x = \frac{\pi}{4} +\pi k$, k∈Z

$x=\frac{\pi }{12} +\frac{\pi k}{3}$, k∈Z

$1\leq \frac{\pi }{12}+\frac{\pi k}{3} \leq 7$

$1\leq \frac{\pi +4\pi k}{12} \leq 7$

$12\leq \pi +4\pi k\leq 84$

$12\leq \pi (1+4k)\leq 84$

$12\leq 3,14(1+4k)\leq 84$

$3,82\leq 1+4k\leq 26,75$

$0,705\leq k\leq 6,4375$

k = 1:  $\frac{\pi }{12} +\frac{\pi }{3} =\frac{5\pi }{12}$

k = 2: $\frac{\pi }{12} +\frac{2\pi }{3} = \frac{3\pi }{4}$

k = 3: $\frac{\pi }{12} +\pi = \frac{13\pi }{12}$

k = 4: $\frac{\pi }{12} +\frac{4\pi }{3} = \frac{17\pi }{12}$

k = 5: $\frac{\pi }{12} +\frac{5\pi }{3} = \frac{21\pi }{12}$

k = 6: $\frac{\pi }{12} +2\pi =\frac{25\pi }{12}$

$-x^2+8x-7=0$

$x=1$

$x=7$

Ответ: 8 корней