(sinx+sin5x)+sin3x=02sin[(x+5x)/2]* cos[(x-5x)/2] + sin3x=02sin(6x/2) * cos(-4x/2)+sin3x=02sin3x * cos2x + sin3x=0sin3x* (2cos2x+1)=0sin3x=0 2cos2x+1=03x=2pi*k, k∈(-∞;+∞) 2cos2x=-1<span>x=2/3 pi*k, k∈(-∞;+∞) cos2x=-1/2</span> 2x=pi- arccos(1/2) 2x= pi- pi/3 +pi*k, k∈(-∞;+∞) 2x= 2pi/3 + pi*k, k∈(-∞;+∞)<span> x= pi/3 + pi/2 *k, k∈(-∞;+∞) </span>
5,3,20,40 Думаю,так я делала своим методом
Решение на фото................................
![1) y= \frac{1}{2} sinx \\ f(x)=\frac{1}{2} sinx \\ f(-x)= \frac{1}{2} sin(-x)=- \frac{1}{2} sinx](https://tex.z-dn.net/?f=1%29+y%3D+%5Cfrac%7B1%7D%7B2%7D+sinx+%5C%5C+f%28x%29%3D%5Cfrac%7B1%7D%7B2%7D+sinx++%5C%5C+f%28-x%29%3D+%5Cfrac%7B1%7D%7B2%7D+sin%28-x%29%3D-+%5Cfrac%7B1%7D%7B2%7D+sinx)
<em>нечетная </em>
![2)y= x^{4} \\ f(x)= x^{4} \\ f(-x)=(-x)^{4} = x^{4}](https://tex.z-dn.net/?f=2%29y%3D+x%5E%7B4%7D++%5C%5C+f%28x%29%3D+x%5E%7B4%7D++%5C%5C+f%28-x%29%3D%28-x%29%5E%7B4%7D+%3D+x%5E%7B4%7D+)
<em>четная </em>
![3) y= \sqrt[3]{x} -cosx \\ f(x)=\sqrt[3]{x} -cosx \\ f(-x)=-\sqrt[3]{x} -cos(-x) =- (\sqrt[3]{x}+cosx) \\](https://tex.z-dn.net/?f=3%29+y%3D+%5Csqrt%5B3%5D%7Bx%7D+-cosx+%5C%5C+f%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D+-cosx++%5C%5C+f%28-x%29%3D-%5Csqrt%5B3%5D%7Bx%7D+-cos%28-x%29+%3D-+%28%5Csqrt%5B3%5D%7Bx%7D%2Bcosx%29+%5C%5C+)
<em>нечетная и нечетная </em>
![4) y=ctgx \\ f(x)=ctgx \\ f(-x)=ctg(-x)=-ctgx](https://tex.z-dn.net/?f=4%29+y%3Dctgx+%5C%5C+f%28x%29%3Dctgx+%5C%5C+f%28-x%29%3Dctg%28-x%29%3D-ctgx)
<em>нечетная </em>
Ответ: 3)
![y= \sqrt[3]{x} -cosx](https://tex.z-dn.net/?f=y%3D+%5Csqrt%5B3%5D%7Bx%7D+-cosx)