Question

решите уравнение 2+3sinx*cosx-2cos2x(все под корнем)=-cosx

Answer 1

$\displaystyle\sqrt{2+3sinxcosx-2cos2x}=-cosx\\2+3sinxcosx-2cos2x=cos^2x\\2sin^2x+2cos^2x+3sinxcosx-2cos^2x+2sin^2x=cos^2x\\4sin^2x+3sinxcosx-cos^2x=0|:cos^2x\neq0;x\neq\frac{\pi}{2}+\pi n\\4tg^2x+3tgx-1=0\\(tgx)_{1,2}=\frac{-3\pm\sqrt{9+16}}{8}=\frac{-3\pm5}{8}\\tgx_1=\frac{1}{4}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ tgx_2=-1\\x_1=arctg\frac{1}{4}+\pi n\ \ \ \ \ x_2=-\frac{\pi}{4}+\pi n;n\in Z$

Проверка найденных корней:

$\displaystyle x_1=arctg\frac{1}{4}\\\sqrt{2+3sinxcosx-2cos2x}=-cosx\\\sqrt{2sin^2x+2cos^2x+3sinxcosx-2cos^2x+2sin^2x}=-cosx\\\sqrt{4sin^2x+3sinxcosx}=-cosx\\\sqrt{4(sin(arctg\frac{1}{4}))^2+3sin(arctg\frac{1}{4})cos(arctg\frac{1}{4})}=-cos(arctg\frac{1}{4})\\\sqrt{4*\frac{\frac{1}{16}}{1+\frac{1}{16}}+3*\frac{\frac{1}{4}}{1+\frac{1}{16}}}=-\frac{1}{\sqrt{1+\frac{1}{16}}}\\\frac{1}{\sqrt{1+\frac{1}{16}}}=-\frac{1}{\sqrt{1+\frac{1}{16}}}$

Получено неверное равенство - корень не подходит.

$\displaystyle x_2=arctg\frac{1}{4}+\pi\\sin(arctg\frac{1}{4}+\pi)=-sin(arctg\frac{1}{4})\\cos(arctg\frac{1}{4}+\pi)=-cos(arctg\frac{1}{4})\\\sqrt{2+3sinxcosx-2cos2x}=-cosx\\\sqrt{2sin^2x+2cos^2x+3sinxcosx-2cos^2x+2sin^2x}=-cosx\\\sqrt{4sin^2x+3sinxcosx}=-cosx\\\sqrt{4(sin(arctg\frac{1}{4}))^2+3sin(arctg\frac{1}{4})cos(arctg\frac{1}{4})}=cos(arctg\frac{1}{4})\\$

$\displaystyle \sqrt{4*\frac{\frac{1}{16}}{1+\frac{1}{16}}+3*\frac{\frac{1}{4}}{1+\frac{1}{16}}}=\frac{1}{\sqrt{1+\frac{1}{16}}}\\\frac{1}{\sqrt{1+\frac{1}{16}}}=\frac{1}{\sqrt{1+\frac{1}{16}}}$

Получено верное равенство. Данный корень подходит

$\displaystyle x_3=-\frac{\pi}{4}\\\sqrt{2+3sinxcosx-2cos2x}=-cosx\\\sqrt{2sin^2x+2cos^2x+3sinxcosx-2cos^2x+2sin^2x}=-cosx\\\sqrt{4sin^2x+3sinxcosx}=-cosx\\\sqrt{4(sin\frac{\pi}{4})^2-3sin\frac{\pi}{4}cos\frac{\pi}{4}}=-cos\frac{\pi}{4}\\\sqrt{2-3*\frac{1}{2}}=-\frac{\sqrt2}{2}\\\frac{\sqrt2}{2}=-\frac{\sqrt2}{2}$

Получено неверное равенство. Данный корень не подходит.

$\displaystyle x_4=-\frac{\pi}{4}+\pi=\frac{3\pi}{4}\\\sqrt{2+3sinxcosx-2cos2x}=-cosx\\\sqrt{2sin^2x+2cos^2x+3sinxcosx-2cos^2x+2sin^2x}=-cosx\\\sqrt{4sin^2x+3sinxcosx}=-cosx\\\sqrt{4(sin\frac{3\pi}{4})^2+3sin(\frac{3\pi}{4})cos(\frac{3\pi}{4})}=-cos\frac{3\pi}{4}\\\sqrt{2-3*\frac{1}{2}}=\frac{\sqrt2}{2}\\\frac{\sqrt2}{2}=\frac{\sqrt2}{2}$

Получено верное равенство. Данный корень подходит.

$\displaystyle x_1=\frac{3\pi}{4}+2\pi n;n\in Z\\x_2=arctg\frac{1}{4}+(2n+1)\pi;n\in Z$