Question

urgent!!!!!!!!!!!!!!!​

Answer 1

Răspuns:

sinxcosx+(cosx)^2=1

sinxcosx=(sinx)^2

sinx(cosx-sinx)=0

Cu solutiile

sinx=0 x= pi

cosx=sinx tgx/1 x=pi+pi/4=5pi/4

Answer 2

$\it x\in\Big(\dfrac{\pi}{2},\ \dfrac{3\pi}{2}\Big)\ \ \ \ (*)\\ \\ \\ sinx+cosx=\dfrac{1}{cosx}\Big |_ {\cdot\ cosx} \Rightarrow sinx\cdot cosx+cos^2x=1\ \Rightarrow sinx\cdotcosx=1-cos^2x\ \Rightarrow \\ \\ \\ \Rightarrow sinx\cdot cosx=sin^2x\ \Rightarrow sinx\cdot cosx-sin^2x=0\ \Rightarrow sinx(cosx-sinx)=0\ \Rightarrow \\ \\ \\ \ \Rightarrow \begin{cases} \it sinx=0 \stackrel{(*)}{\Longrightarrow}x=\pi\\ \\ \it cosx-sinx=0\ \stackrel{(*)}{\Longrightarrow}\ x=\dfrac{5\pi}{4}\end{cases}$