\begin{equation}
\begin{split}
6\cos^2x+5\sin x &= 7\\
6(1-\sin^2x)+5\sin x &= 7\\
\text{Legyen: }\color{darkblue}{y:=\sin x}\\
\color{darkblue}{6(1-y^2)+5y} &\color{dargblue}{=} \color{darkblue}{7}\\
\color{darkblue}{6-6y^2+5y-7} &\color{dargblue}{=} \color{darkblue}{0}\\
\color{darkblue}{-6y^2+5y-1} &\color{dargblue}{=} \color{darkblue}{0}\\
\color{darkblue}{6y^2-5y+1} &\color{dargblue}{=} \color{darkblue}{0}
\end{split}
\end{equation}

\begin{equation} \begin{split} \color{darkblue}{y_1} &\color{darkblue}{=} \color{darkblue}{\tfrac13}\\ \sin x &= \tfrac13\\\\ \color{darkred}{x_1} &\color{darkred}{=} \color{darkred}{0{,}3398+k\cdot2\pi}\\ \color{darkred}{x_2} &\color{darkred}{=} \color{darkred}{2{,}802+k\cdot2\pi};\hphantom{00}\color{darkred}{k\in\mathbb Z} \end{split} \end{equation}

\begin{equation} \begin{split} \color{darkblue}{y_2} &\color{darkblue}{=} \color{darkblue}{\tfrac12}\\ \sin x &= \tfrac12\\\\ \color{darkred}{x_3} &\color{darkred}{=} \color{darkred}{\tfrac{\pi}6+k\cdot2\pi}\\ \color{darkred}{x_4} &\color{darkred}{=} \color{darkred}{\tfrac{5}6\pi+k\cdot2\pi};\hphantom{00}\color{darkred}{k\in\mathbb Z} \end{split} \end{equation}

Kik azok, akiknek a koszinusza kisebb, mint \(-\frac{\sqrt{2}}2\) ?

 
Válasz: \(\frac34\pi+k\cdot 2\pi<2x<\frac54\pi+k\cdot2\pi\)
 
2-vel osztva: \(\frac38\pi+k\cdot\pi<x<\frac58\pi+k\cdot\pi\)
\[
\color{darkred}{\textbf{M.:}\hphantom{00}\mathbf{\left]\tfrac38\pi+k\cdot\pi;\tfrac58\pi+k\cdot\pi\right[;\hphantom{000}k\in\mathbb Z}}
\]

\begin{equation}
\begin{split}
4\sin x -3\cos x &= 2\\
\color{darkblue}{y:=\tfrac{x}2}\\
\color{darkblue}{4\sin 2y+3\cos 2y} &\color{darkblue}{=} \color{darkblue}{2}\\
\color{darkblue}{4\cdot2\sin y\cos y + 3(\cos^2y-\sin^2y)} &\color{darkblue}{=} \color{darkblue}{2}\\
\color{darkblue}{8\sin y\cos y+3\cos^2y-3\sin^2y} &\color{darkblue}{=} \color{darkblue}{2}\\
\text{Mivel }\sin^2y+\cos^2y=1,\\
\text{azért }2=2\sin^2y+2\cos^2y\\
\color{darkblue}{8\sin y\cos y+3\cos^2y-3\sin^2y} &\color{darkblue}{=} \color{darkblue}{2\sin^2y+2\cos^2y}\\
\color{darkblue}{-5\sin^2y+8\sin y\cos y+\cos^2y} &\color{darkblue}{=} \color{darkblue}{0}
\hphantom{0000}\color{darkblue}{\big/\,:\cos^2x}\hphantom{00}\color{darkgreen}{\text{Felt.: }\cos x\ne0}\\
\color{darkblue}{-5\text{tg}^2y+8\text{tg}\,y+1} &\color{darkblue}{=} \color{darkblue}{0}\\
\color{brown}{T:=\text{tg}\,y}\\
\color{brown}{-5T^2+8T+1} &\color{brown}{=} \color{brown}{0}
\end{split}
\end{equation}

\begin{equation} \begin{split} \color{brown}{T_1} &\color{brown}{=} \color{brown}{1{,}7165}\\ \color{darkblue}{\text{tg}\,y_1} &\color{darkblue}{=} \color{darkblue}{1{,}7165}\\ \color{darkblue}{y_1} &\color{darkblue}{=} \color{darkblue}{59{,}78^\circ+k\cdot180^\circ}\\ \tfrac{x_1}2 &= 59{,}78^\circ + k\cdot180^\circ\\ x_1 &= 119{,}56+k\cdot360^\circ \end{split} \end{equation}

\begin{equation} \begin{split} \color{brown}{T_2} &\color{brown}{=} \color{brown}{-0{,}1165}\\ \color{darkblue}{\text{tg}\,y_2} &\color{darkblue}{=} \color{darkblue}{-0{,}1165}\\ \color{darkblue}{y_2} &\color{darkblue}{=} \color{darkblue}{-6{,}645^\circ+k\cdot180^\circ}\\ \tfrac{x_2}2 &= -6{,}645^\circ + k\cdot180^\circ\\ x_2 &= -13{,}29+k\cdot360^\circ \end{split} \end{equation}

Feltételvizsgálat: Ha \(\cos y=0\), akkor \(\sin y=1\) vagy \(\sin y=-1\).
- Ha \(\sin y=1\), akkor behelyettesítve (a mûvelet elõtti állapotba): \(-5\cdot1+8\cdot1\cdot0+0=-5\ne2\).
- Ha \(\sin y=-1\), akkor behelyettesítve (a mûvelet elõtti állapotba): \(-5\cdot1+8\cdot(-1)\cdot0+0=-5\ne2\).
Tehát a feltételezésünkkel nem vesztettünk megoldást.
  
A \(\left[0^\circ;360^\circ\right]\) intervallumba esõ megoldások:
\[
\color{darkred}{\mathbf{x_1=119{,}56^\circ\hphantom{00000000000000000000}x_2=346{,}71^\circ}}
\]