Ответы
Ответ дал:
0
Думаю, можно разобраться...
Приложения:
Ответ дал:
0
a)![16cos^4x-24cos^2x+9=0;\
cos x=t, -1leq xleq1;\
16t^4-24t^2+9=0;\
(4t^2)^2-2cdot4t^2cdot3+3^2=0;\
(4t^2-3)=0;\
4t^2-3=0;\
4t^2=3;\
t^2=frac{3}{4};\
t=pmfrac{sqrt3}{2};\](https://tex.z-dn.net/?f=16cos%5E4x-24cos%5E2x%2B9%3D0%3B%5C%0Acos+x%3Dt%2C++++-1leq+xleq1%3B%5C%0A16t%5E4-24t%5E2%2B9%3D0%3B%5C%0A%284t%5E2%29%5E2-2cdot4t%5E2cdot3%2B3%5E2%3D0%3B%5C%0A%284t%5E2-3%29%3D0%3B%5C%0A4t%5E2-3%3D0%3B%5C%0A4t%5E2%3D3%3B%5C%0At%5E2%3Dfrac%7B3%7D%7B4%7D%3B%5C%0At%3Dpmfrac%7Bsqrt3%7D%7B2%7D%3B%5C "16cos^4x-24cos^2x+9=0;\
cos x=t, -1leq xleq1;\
16t^4-24t^2+9=0;\
(4t^2)^2-2cdot4t^2cdot3+3^2=0;\
(4t^2-3)=0;\
4t^2-3=0;\
4t^2=3;\
t^2=frac{3}{4};\
t=pmfrac{sqrt3}{2};\")
![cos x=pmfrac{sqrt3}{2};\
cos x=frac{sqrt3}{2};\
x=pmarccosfrac{sqrt3}{2}+2pi n, nin Z;\
x=pmfrac{pi}{6}+2pi n, nin Z;\
cos x=-frac{sqrt3}{2};\
x=pmarccos(-frac{sqrt3}{2})+2pi k, k in Z;\
x=pmfrac{5pi}{6}+2pi k;\
left[ {{x=pmfrac{pi}{6}+2pi n,} atop {x=pipmfrac{pi}{6}+2pi k}} right. n,kin Z](https://tex.z-dn.net/?f=cos+x%3Dpmfrac%7Bsqrt3%7D%7B2%7D%3B%5C%0Acos+x%3Dfrac%7Bsqrt3%7D%7B2%7D%3B%5C%0Ax%3Dpmarccosfrac%7Bsqrt3%7D%7B2%7D%2B2pi+n%2C+nin+Z%3B%5C%0Ax%3Dpmfrac%7Bpi%7D%7B6%7D%2B2pi+n%2C+nin+Z%3B%5C%0Acos+x%3D-frac%7Bsqrt3%7D%7B2%7D%3B%5C%0Ax%3Dpmarccos%28-frac%7Bsqrt3%7D%7B2%7D%29%2B2pi+k%2C+k+in+Z%3B%5C%0Ax%3Dpmfrac%7B5pi%7D%7B6%7D%2B2pi+k%3B%5C%0A+left%5B+%7B%7Bx%3Dpmfrac%7Bpi%7D%7B6%7D%2B2pi+n%2C%7D+atop+%7Bx%3Dpipmfrac%7Bpi%7D%7B6%7D%2B2pi+k%7D%7D+right.++++n%2Ckin+Z "cos x=pmfrac{sqrt3}{2};\
cos x=frac{sqrt3}{2};\
x=pmarccosfrac{sqrt3}{2}+2pi n, nin Z;\
x=pmfrac{pi}{6}+2pi n, nin Z;\
cos x=-frac{sqrt3}{2};\
x=pmarccos(-frac{sqrt3}{2})+2pi k, k in Z;\
x=pmfrac{5pi}{6}+2pi k;\
left[ {{x=pmfrac{pi}{6}+2pi n,} atop {x=pipmfrac{pi}{6}+2pi k}} right. n,kin Z")
б)
![xin [2pi,3pi];\
2pi+frac{pi}{6}=frac{12pi+pi}{6}=frac{13pi}{6};\
2pi+pi-frac{pi}{6}=3pi-frac{pi}{6}=frac{18pi-pi}{6}=frac{17pi}{6};\
\
\
\
x=frac{13pi}{6};\
x=frac{17pi}{6};\](https://tex.z-dn.net/?f=xin+%5B2pi%2C3pi%5D%3B%5C%0A2pi%2Bfrac%7Bpi%7D%7B6%7D%3Dfrac%7B12pi%2Bpi%7D%7B6%7D%3Dfrac%7B13pi%7D%7B6%7D%3B%5C%0A2pi%2Bpi-frac%7Bpi%7D%7B6%7D%3D3pi-frac%7Bpi%7D%7B6%7D%3Dfrac%7B18pi-pi%7D%7B6%7D%3Dfrac%7B17pi%7D%7B6%7D%3B%5C%0A%5C%0A%5C%0A%5C%0Ax%3Dfrac%7B13pi%7D%7B6%7D%3B%5C%0Ax%3Dfrac%7B17pi%7D%7B6%7D%3B%5C "xin [2pi,3pi];\
2pi+frac{pi}{6}=frac{12pi+pi}{6}=frac{13pi}{6};\
2pi+pi-frac{pi}{6}=3pi-frac{pi}{6}=frac{18pi-pi}{6}=frac{17pi}{6};\
\
\
\
x=frac{13pi}{6};\
x=frac{17pi}{6};\")
б)
Приложения:
Похожие вопросы
7 лет назад
10 лет назад
10 лет назад