Ответ во вложении. Если что-то непонятно, пиши.
Ответ:
Объяснение:
,
где t - любая переменная или функция.
![f(\underbrace {\varphi (x)}_{t})=\underbrace {2^{\varphi (x)}}_{2^{t}}=2^{x^4}\\\\\varphi (\underbrace {f(x)}_{t})=\underbrace {cos(f(x))}_{cost}=cos(2^{x})\\\\f(\, \underbrace {g(\varphi (x))}_{t}\, )=2^{g(\varphi (x))}=2^{cos(\varphi (x))}=2^{cos(x^4)}](https://tex.z-dn.net/?f=f%28%5Cunderbrace%20%7B%5Cvarphi%20%28x%29%7D_%7Bt%7D%29%3D%5Cunderbrace%20%7B2%5E%7B%5Cvarphi%20%28x%29%7D%7D_%7B2%5E%7Bt%7D%7D%3D2%5E%7Bx%5E4%7D%5C%5C%5C%5C%5Cvarphi%20%28%5Cunderbrace%20%7Bf%28x%29%7D_%7Bt%7D%29%3D%5Cunderbrace%20%7Bcos%28f%28x%29%29%7D_%7Bcost%7D%3Dcos%282%5E%7Bx%7D%29%5C%5C%5C%5Cf%28%5C%2C%20%5Cunderbrace%20%7Bg%28%5Cvarphi%20%28x%29%29%7D_%7Bt%7D%5C%2C%20%29%3D2%5E%7Bg%28%5Cvarphi%20%28x%29%29%7D%3D2%5E%7Bcos%28%5Cvarphi%20%28x%29%29%7D%3D2%5E%7Bcos%28x%5E4%29%7D)
![2\; var.\; \; 1)\; \;x=81\; ,\; \; f(x)=(log_3\sqrt{x})^4=(log_3\sqrt{81})^4=(log_39)^4=\\\\=(log_33^2)^4=(2\, log_33)^4=(2\cdot 1)^4=2^4=16\\\\\\x=\frac{\pi ^2}{4}\; ,\; \; y=sin(cos\sqrt{x}\, )=sin(cos\sqrt{\frac{\pi ^2}{4}})=sin(cos\frac{\pi}{2})=sin0=0](https://tex.z-dn.net/?f=2%5C%3B%20var.%5C%3B%20%5C%3B%201%29%5C%3B%20%5C%3Bx%3D81%5C%3B%20%2C%5C%3B%20%5C%3B%20%20f%28x%29%3D%28log_3%5Csqrt%7Bx%7D%29%5E4%3D%28log_3%5Csqrt%7B81%7D%29%5E4%3D%28log_39%29%5E4%3D%5C%5C%5C%5C%3D%28log_33%5E2%29%5E4%3D%282%5C%2C%20log_33%29%5E4%3D%282%5Ccdot%201%29%5E4%3D2%5E4%3D16%5C%5C%5C%5C%5C%5Cx%3D%5Cfrac%7B%5Cpi%20%5E2%7D%7B4%7D%5C%3B%20%2C%5C%3B%20%5C%3B%20y%3Dsin%28cos%5Csqrt%7Bx%7D%5C%2C%20%29%3Dsin%28cos%5Csqrt%7B%5Cfrac%7B%5Cpi%20%5E2%7D%7B4%7D%7D%29%3Dsin%28cos%5Cfrac%7B%5Cpi%7D%7B2%7D%29%3Dsin0%3D0)
![3)\; \; g(x)=\sqrt[3]{x}\; \; ,\; \; \varphi (x)=5^{x}\; \; ,\; \; f(x)=sinx\\\\f(\varphi (x))=sin(\varphi (x))=sin(5^{x})\\\\\varphi (f(x))=5^{f(x)}=5^{sinx}\\\\f(g(\varphi (x)))=sin(g(\varphi (x))=sin(\sqrt[3]{\varphi (x)}\, )=sin(\sqrt[3]{5^{x}})](https://tex.z-dn.net/?f=3%29%5C%3B%20%5C%3B%20g%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D%5C%3B%20%5C%3B%20%2C%5C%3B%20%5C%3B%20%5Cvarphi%20%28x%29%3D5%5E%7Bx%7D%5C%3B%20%5C%3B%20%2C%5C%3B%20%5C%3B%20f%28x%29%3Dsinx%5C%5C%5C%5Cf%28%5Cvarphi%20%28x%29%29%3Dsin%28%5Cvarphi%20%28x%29%29%3Dsin%285%5E%7Bx%7D%29%5C%5C%5C%5C%5Cvarphi%20%28f%28x%29%29%3D5%5E%7Bf%28x%29%7D%3D5%5E%7Bsinx%7D%5C%5C%5C%5Cf%28g%28%5Cvarphi%20%28x%29%29%29%3Dsin%28g%28%5Cvarphi%20%28x%29%29%3Dsin%28%5Csqrt%5B3%5D%7B%5Cvarphi%20%28x%29%7D%5C%2C%20%29%3Dsin%28%5Csqrt%5B3%5D%7B5%5E%7Bx%7D%7D%29)
1) Посмотри, какой приём при решении таких уравнений есть.
<span>Обозначим </span>tg x/2 = t, тогда Cos x = (1 - t²)/(1 + t²) и
Sin x = 2t /(1 + t²)
Сделаем замену в нашем уравнении.
5(1 - t²)/(1 + t²) + 12·2t/(1 + t²) = 13 | · (1 + t²)≠0
5(1 - t²) +24 t = 13 + 13 t²
18 t² - 24 t +8 = 0
9t² - 12 t +4 = 0
t = 2/3
tg x/2 = 2/3
х/2 = arc tg 2/3 + πк, где к∈Z
x = 2 arc tg 2/3 + 2πк, где к ∈Z
2)3 Cos x - 2 ·2sin x Cos x = 0
Cos x(3 - 4Sin x) = 0
Cos x = 0 или 3 - 4 Sin x = 0
x = π/2 + πr, где к ∈Z<span> 4Sin x = 3</span>
Sin x = 3/4
x = (-1)^k arcSin 3/4 + кπ, где к ∈z