Question

Найдите корни которые принадлежат интервалу [π;2.5π]
Просьба, можно пожалуйста подробно показать отбор корней​

Answer 1

Ответ: 1,25π

Объяснение: на фото

Answer 2

$2Sin^{2}x=4+3\sqrt{2}Sin\Big(\dfrac{\pi }{2} -x\Big)\\\\2(1-Cos^{2}x)=4+3\sqrt{2}Cosx\\\\2-2Cos^{2}x-4-3\sqrt{2}Cosx=0\\\\2Cos^{2}x+3\sqrt{2}Cosx+2=0\\\\Cosx=m \ , \ -1\leq m\leq 1\\\\2m^{2} +3\sqrt{2}m+2=0\\\\D=(3\sqrt{2})^{2} -4\cdot 2\cdot 2=18-16=2=(\sqrt{2})^{2} \\\\m_{1}=\dfrac{-3\sqrt{2}+\sqrt{2}}{4}=-\dfrac{2\sqrt{2} }{4}=-\dfrac{\sqrt{2} }{2} \\\\m_{2} =\dfrac{-3\sqrt{2}-\sqrt{2}}{4} =-\dfrac{4\sqrt{2} }{4}=-\sqrt{2}<-1-neyd$

$Cosx=-\dfrac{\sqrt{2} }{2} \\\\x=\pm\dfrac{3\pi }{4}+2\pi n,n\in Z\\\\1)x=-\dfrac{3\pi }{4}+2\pi n,n\in Z\\\\\pi \leq-\dfrac{3\pi }{4}+2\pi n\leq2,5\pi \\\\\pi+\dfrac{3\pi }{4} \leq 2\pi n\leq\dfrac{5\pi }{2}+\dfrac{3\pi }{4}\\\\\dfrac{5\pi }{4}\leq 2\pi n\leq \dfrac{13\pi }{4}\\\\\dfrac{5}{4}\leq 2n\leq \dfrac{13}{4}$

$\dfrac{5}{8}\leq n\leq 1\dfrac{5}{8} \\\\n=1 \ \Rightarrow \ x=-\dfrac{3\pi }{4}+2\pi=-\dfrac{3\pi }{4}+\dfrac{8\pi }{4}=\dfrac{5\pi }{4}=\boxed{1,25\pi}$

$2)x=\dfrac{3\pi }{4}+2\pi n,n\in Z\\\\\pi \leq\dfrac{3\pi }{4}+2\pi n\leq2,5\pi \\\\\pi-\dfrac{3\pi }{4} \leq 2\pi n\leq\dfrac{5\pi }{2}-\dfrac{3\pi }{4}\\\\\dfrac{\pi }{4}\leq 2\pi n\leq \dfrac{7\pi }{4}\\\\\dfrac{1}{4}\leq 2n\leq \dfrac{7}{4}$

$\dfrac{1}{8} \leq n\leq \frac{7}{8}$

Здесь нет целых значений n , значит корней больше нет .

Ответ : 1,25π