Răspuns:
2
Explicație pas cu pas:
[tex]\sin x + \cos {4x} = 2\iff \begin{cases}\sin{x} = 1\\\cos{4x}=1\end{cases}\\\\[/tex]
[tex]Fie\:z = 2x\\\\\cos{4x} = \cos{2z} = 2cos^2(z) - 1\\ cos(z) = \cos{2x} = 1 - 2sin^2(x)\\\\\cos{4x} = 2(1 - 2sin^2(x))^2 - 1 = 1\Big{|+1} \\\\2(1 - 2sin^2(x))^2 = 2\Big{|\div 2}\\\\(1-2sin^2(x))^2 = 1\\\\1 - 4sin^2(x) + 4sin^4(x) = 1\Big{|-1+4sin^2(x)}\\\\4sin^4(x) = 4sin^2(x)\Big{|\div 4}\\\\(sin^2(x))^2 = sin^2(x)\Big{|\div sin^2(x)}\\\\sin^2(x) = 1\\\\ \sin{x} = \pm 1[/tex]
[tex]\begin{cases}\sin{x} = 1\\\cos{4x}=1\end{cases} \iff \begin{cases}\sin{x} = 1\\\sin{x}=\pm 1\end{cases} \\\\\implies \sin{x} = 1\implies x \in \Big\{(4k + 1)\frac{\pi}{2}\Big | k \in \mathbb{Z}\Big\}[/tex]
Acum, daca intersectam aceasta multime cu domeniul de definitie al functiei, obtinem:
[tex]S = \Big\{\frac{\pi}{2}, 2\pi + \frac{\pi}{2}\Big\}\Rightarrow card\: S = 2[/tex]
