Question

Determinați x aparține (0,pi), știind ca sin2x+6 cosx-sinx-3=0

Answer 1

$\it sin2x+6cosx-sinx-3=0 \Rightarrow 2sinxcosx+6cosx-sinx-3=0 \Rightarrow \\ \\ \Rightarrow sinx(2cosx-1)+3(2cosx-1)=0 \Rightarrow (2cosx-1)(sinx+3)=0 \Rightarrow \\ \\ \Rightarrow 2cosx-1=0 \Rightarrow cosx=\dfrac{1}{2}\ \stackrel{x\in(0,\ \pi)}{\Longrightarrow}\ \ x=\dfrac{\pi}{3}$