\begin{equation}
\begin{split}
\textbf{A)}\hphantom{000}\cos3x &= \overbrace{\cos(2x+x) = \color{darkblue}{\cos 2x}\cos x -\color{darkgreen}{\sin2x}\sin x}^{\text{Alkalmazva: }\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta} = (\color{darkblue}{\cos^2x-\sin^2x})\cos x - \color{darkgreen}{2\sin x\cos x}\sin x = \\
&= \cos^3x - \sin^2x\cos x -2\sin^2x\cos x = \color{darkred}{\mathbf{\cos^3x - 3\sin^2x\cos x}}
\\\\
\textbf{B)}\hphantom{000}\sin4x &= \sin (2\cdot2x) = 2\color{darkblue}{\sin2x}\cdot\color{darkgreen}{\cos2x} = 2\cdot\color{darkblue}{2\sin x\cos x}\cdot(\color{darkgreen}{\cos^2x-\sin^2x}) =\\
&= \color{darkred}{\mathbf{4\sin x\cos^3x - 4\sin^3x\cos x}}
\end{split}
\end{equation}

\begin{equation}
\begin{split}
\textbf{A)}\hphantom{000}\color{darkblue}{\cos(x+y)}+\color{darkgreen}{\cos(x-y)} &= \color{darkblue}{\cos x\cos y-\sin x\sin y}+\color{darkgreen}{\cos x\cos y +\sin x\sin y} =\\
&= \color{darkred}{\mathbf{2\cos x\cos y}}
\\\\
\textbf{B)}\hphantom{000}\color{darkblue}{\sin(x+y)}-\color{darkgreen}{\sin(x-y)} &=
\color{darkblue}{\sin x\cos y+\cos x\sin y}-\color{darkgreen}{(\sin x\cos y-\cos x\sin y)}=\\
&= \sin x\cos y+\cos x\sin y -\sin x\cos y +\cos x\sin y=\\
&= \color{darkred}{\mathbf{2\cos x\sin y}}
\end{split}
\end{equation}

A)
\begin{equation}
\begin{split}
4\cos x &= \tfrac1{\sin x}\hphantom{000000}&\big/\cdot\sin x\hphantom{000000000000000000000000000000}&\color{red}{\textbf{K.: }\sin x\ne0}\\
4\sin x\cos x &= 1 &\big/\,2\sin x\cos x=\sin2x&\color{red}{x\ne k\pi;\,k\in\mathbb Z}\\
2\sin2x &= 1 &\big/\,:2\\
\sin 2x &=\tfrac12\\\\
2x_1 &= \tfrac{\pi}6 + k\cdot 2\pi\\
2x_2 &= \tfrac56\pi + k\cdot 2\pi\\\\
\color{darkred}{\mathbf{x_1}} &\color{darkred}{=} \color{darkred}{\mathbf{\tfrac{\pi}{12} + k\cdot \pi}}\\
\color{darkred}{\mathbf{x_2}} &\color{darkred}{=} \color{darkred}{\mathbf{\tfrac{5}{12}\pi + k\cdot \pi}} &\hphantom{0000}\color{darkred}{k\in\mathbb Z}
\end{split}
\end{equation}A kikötésnek megfelelnek, és csupa ekvivalens étalakítást végeztünk.
 
 
B)
\begin{equation}
\begin{split}
\sin x +\cos x &= 1\\
\text{Alkalmazzuk: }x=2y\\
\sin 2y + \cos 2y &= 1\\
2\sin y\cos y + \cos^2y-\sin^2y &= 1\\
2\sin y\cos y + \cos^2y-\sin^2y &= \sin^2y + \cos^2y\\
2\sin y\cos y -2\sin^2y &= 0\\
2\sin y(\cos y-\sin y) &= 0\\\\
\text{1. lehetõség: }\\
\sin y &= 0\\
y_1 &= k\cdot \pi\\
\tfrac{x_1}2 &= k\cdot\pi\\
\color{darkred}{\mathbf{x_1}} &\color{darkred}{=} \color{darkred}{\mathbf{k\cdot 2\pi;}\hphantom{000}k\in\mathbb Z}\\\\
\text{2. lehetõség: }\\
\cos y-\sin y &= 0\\
\cos y &= \sin y &\big/\,:\cos y\hphantom{00000000}\color{green}{\text{F.: }\cos y\ne 0}\\
1 &= \text{tg}\,y\\
y_2 &= \tfrac\pi4+k\cdot\pi\\
\tfrac{x_2}2 &= \tfrac\pi4+k\cdot\pi\\
\color{darkred}{\mathbf{x_2}} &\color{darkred}{=} \color{darkred}{\mathbf{\tfrac\pi2+k\cdot 2\pi;}\hphantom{000}k\in\mathbb Z}
\end{split}
\end{equation}
Feltételvizsgálat: Megoldja-e a \(\cos y=\sin y\) egyenletet, ha \(\cos y=0\)?
Válasz: NEM, mert ha \(\cos y=0\), akkor \(\sin y=\pm1\), tehát \(\cos y=\sin y\) nem teljesül.
 
Csupa ekvivalens átalakítást végeztünk, ellenõrizni nem kell, mindkét gyök jó.