Предмет: Алгебра,
автор: Rorsherh
Как решаются такие уроавнения? Помогите!
Приложения:
![](https://files.topotvet.com/i/ae0/ae0eebc912d02656e8a11632cf975891.jpg)
Ответы
Автор ответа:
1
во первых
![\frac{n}{cos^2x} =n(1+tg^2x) \frac{n}{cos^2x} =n(1+tg^2x)](https://tex.z-dn.net/?f=+%5Cfrac%7Bn%7D%7Bcos%5E2x%7D+%3Dn%281%2Btg%5E2x%29)
![6sin^2x+sinxcosx-cos^2x=2 \ \ (:cos^2x \neq 0) \\ 6tg^2x+tgx-1=2(tg^2x+1) \\ 6tg^2x+tgx-1=2tg^2x+2 \\ 4tg^2x+tgx-3=0 \\ \\ tgx=a \\ \\ 4a^2+a-3=0 \\ D=1+48=7^2 \\ a_1= \frac{-1-7}{8}=-1 \\ \\ a_2= \frac{-1+7}{8}= \frac{3}{4} \\ \\ tgx=-1 \\ x=- \frac{ \pi }{4} + \pi k, k\in Z \\ \\ tgx= \frac{3}{4} \\ x=arctg( \frac{3}{4} )+ \pi k, k \in Z 6sin^2x+sinxcosx-cos^2x=2 \ \ (:cos^2x \neq 0) \\ 6tg^2x+tgx-1=2(tg^2x+1) \\ 6tg^2x+tgx-1=2tg^2x+2 \\ 4tg^2x+tgx-3=0 \\ \\ tgx=a \\ \\ 4a^2+a-3=0 \\ D=1+48=7^2 \\ a_1= \frac{-1-7}{8}=-1 \\ \\ a_2= \frac{-1+7}{8}= \frac{3}{4} \\ \\ tgx=-1 \\ x=- \frac{ \pi }{4} + \pi k, k\in Z \\ \\ tgx= \frac{3}{4} \\ x=arctg( \frac{3}{4} )+ \pi k, k \in Z](https://tex.z-dn.net/?f=6sin%5E2x%2Bsinxcosx-cos%5E2x%3D2+%5C+%5C+%28%3Acos%5E2x++%5Cneq+0%29+%5C%5C+6tg%5E2x%2Btgx-1%3D2%28tg%5E2x%2B1%29+%5C%5C+6tg%5E2x%2Btgx-1%3D2tg%5E2x%2B2+%5C%5C+4tg%5E2x%2Btgx-3%3D0+%5C%5C++%5C%5C+tgx%3Da+%5C%5C++%5C%5C+4a%5E2%2Ba-3%3D0+%5C%5C+D%3D1%2B48%3D7%5E2+%5C%5C+a_1%3D+%5Cfrac%7B-1-7%7D%7B8%7D%3D-1+%5C%5C+%5C%5C++a_2%3D+%5Cfrac%7B-1%2B7%7D%7B8%7D%3D+++%5Cfrac%7B3%7D%7B4%7D++%5C%5C++%5C%5C+tgx%3D-1+%5C%5C+x%3D-+%5Cfrac%7B+%5Cpi+%7D%7B4%7D+%2B++%5Cpi+k%2C+k%5Cin+Z+%5C%5C++%5C%5C+tgx%3D+%5Cfrac%7B3%7D%7B4%7D++%5C%5C+x%3Darctg%28+%5Cfrac%7B3%7D%7B4%7D+%29%2B+%5Cpi+k%2C+k+%5Cin+Z)
![3sin^2x+4sinxcosx-3cos^2x-2=0 \ \ (:cos^2x \neq 0 ) \\ 3tg^2x+4tgx-3-2(1+tg^2x)=0 \\ 3tg^2x+4tgx-3-2-2tg^2x=0 \\ tg^2x+4tgx-5=0 \\ D=16+20=6^2 \\ tgx_1= \frac{-4+6}{2} =1 \\ \\ tgx_2= \frac{-4-6}{2}=-5 \\ \\ x_1= \frac{ \pi }{4} + \pi k, k \in Z \\ x_2=-arctg(5)+ \pi k, k\in Z 3sin^2x+4sinxcosx-3cos^2x-2=0 \ \ (:cos^2x \neq 0 ) \\ 3tg^2x+4tgx-3-2(1+tg^2x)=0 \\ 3tg^2x+4tgx-3-2-2tg^2x=0 \\ tg^2x+4tgx-5=0 \\ D=16+20=6^2 \\ tgx_1= \frac{-4+6}{2} =1 \\ \\ tgx_2= \frac{-4-6}{2}=-5 \\ \\ x_1= \frac{ \pi }{4} + \pi k, k \in Z \\ x_2=-arctg(5)+ \pi k, k\in Z](https://tex.z-dn.net/?f=3sin%5E2x%2B4sinxcosx-3cos%5E2x-2%3D0+%5C+%5C+%28%3Acos%5E2x+%5Cneq+0+%29+%5C%5C+3tg%5E2x%2B4tgx-3-2%281%2Btg%5E2x%29%3D0+%5C%5C+3tg%5E2x%2B4tgx-3-2-2tg%5E2x%3D0+%5C%5C+tg%5E2x%2B4tgx-5%3D0+%5C%5C+D%3D16%2B20%3D6%5E2+%5C%5C+tgx_1%3D+%5Cfrac%7B-4%2B6%7D%7B2%7D+%3D1+%5C%5C+%5C%5C++tgx_2%3D+%5Cfrac%7B-4-6%7D%7B2%7D%3D-5++%5C%5C++%5C%5C+x_1%3D+%5Cfrac%7B+%5Cpi+%7D%7B4%7D+%2B+%5Cpi+k%2C+k+%5Cin+Z+%5C%5C+x_2%3D-arctg%285%29%2B+%5Cpi+k%2C+k%5Cin+Z)
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