Question

find the general solution of 2 sin²x + (root 3)cos x + 1=0

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{equation}\mathrm{is}, \\ 2{\sin }^2\mathrm{x}+\sqrt{3}\cos \mathrm{x}+1=0 \\ \Rightarrow 2\left(1-{\cos }^2\mathrm{x}\right)+\sqrt{3}\cos \mathrm{x}+1=0 \\ \Rightarrow 2-2{\cos }^2\mathrm{x}+\sqrt{3}\cos \mathrm{x}+1=0 \\ \Rightarrow 2{\cos }^2\mathrm{x}-\sqrt{3}\cos \mathrm{x}-3=0 \\ \Rightarrow 2{\cos }^2\mathrm{x}-2\sqrt{3}\cos \mathrm{x}+\sqrt{3}\cos \mathrm{x}-3=0 \\ \Rightarrow 2\cos \mathrm{x}\left(\cos \mathrm{x}-\sqrt{3}\right)+\sqrt{3}\left(\cos \mathrm{x}-\sqrt{3}\right)=0 \\ \Rightarrow \left(\cos \mathrm{x}-\sqrt{3}\right)\left(2\cos \mathrm{x}+\sqrt{3}\right)=0 \\ \Rightarrow 2\cos \mathrm{x}+\sqrt{3}=0\mathrm{or}\cos \mathrm{x}-\sqrt{3}=0 \\ \Rightarrow \cos \mathrm{x}=-\frac{\sqrt{3}}{2}\mathrm{or}\cos \mathrm{x}=\sqrt{3}\mathbf{}\left(\mathbf{rejected}\right) \\ \Rightarrow \cos \mathrm{x}=-\frac{\sqrt{3}}{2} \\ \Rightarrow \cos \mathrm{x}=-\cos \left(\frac{\mathrm{\pi }}{6}\right) \\ \Rightarrow \cos \mathrm{x}=\cos \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right) \\ \Rightarrow \cos \mathrm{x}=\cos \left(\frac{5\mathrm{\pi }}{6}\right) \\ \Rightarrow \mathrm{x}=2\mathrm{n\pi }\pm \frac{5\mathrm{\pi }}{6};\mathrm{where}\mathrm{n}\in \mathrm{Z} \end{gathered}$$