Question

$Sa~se~rezolve~exercitiul~sinx+cosx=1~cu~metoda~sistemului~de~ec.$

Answer 1

$\underset{\left\| \begin{array}{ll} Notam \sin x = a \\ Notam \cos x = b \end{array} \right }{\underbrace{\left\{ \begin{array}{ll} \sin x +\cos x = 1 \\ \sin^2 x+\cos^2 x = 1 \end{array} \right }} \Rightarrow \left\{ \begin{array}{ll} a+b = 0 \\ a^2+b^2 = 0 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} a+b=1 \\ (a+b)^2-2ab = 1 \end{array} \right \Rightarrow$$\Rightarrow \left\{ \begin{array}{ll} a+b=1 \\ 1-2ab = 1 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} a+b = 1 \\ 2ab = 0 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} a+b = 1 \\ ab = 0 \end{array} \right| \\ \\ t^2-St+P = 0 \Rightarrow t^2-t+0 =0 \Rightarrow t(t-1) = 0$$\Rightarrow \left\{ \begin{array}{ll} a = 0 \\ b = 1 \end{array} \right sau \left \begin{array}{ll} a=1 \\ b=0 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} \sin x= 0 \\ \cos x = 1 \end{array} \right sau \left \begin{array}{ll} \sin x=1 \\ \cos x=0 \end{array} \right \Rightarrow$$\Rightarrow \left\{ \begin{array}{ll} x = 0\\ \\ x = 0\end{array} \right sau \left \begin{array}{ll} x = \dfrac{\pi}{2} \\ \\ x = \dfrac{\pi}{2} \end{array} \right| \Rightarrow \boxed{\boxed{S = \Big\{0,\dfrac{\pi}{2}\Big\}}}$