Given the trigonometry equation below
[tex](2\cos x+1)(2\sin x+\sqrt[]{3})=0[/tex]If two numbers multiplies themselves and the result is zero, it implies that one of them is zero or both are zero
For instance:
[tex]\begin{gathered} \text{If} \\ a\times b=0 \\ a=0\text{ or }b=0 \end{gathered}[/tex]Thus,
[tex]\begin{gathered} 2\cos x+1=0 \\ OR \\ 2\sin x+\sqrt[]{3}=0 \end{gathered}[/tex][tex]\begin{gathered} 2\cos x+1=0 \\ \Rightarrow2\cos x=-1 \\ \Rightarrow\frac{2\cos x}{2}=\frac{-1}{2} \\ \therefore cosx=-\frac{1}{2} \\ x=\cos ^{-1}-\frac{1}{2} \\ \text{Values of x}\Rightarrow120^0,240^0 \\ In\text{ radian} \end{gathered}[/tex][tex]\begin{gathered} 120^0\times\frac{\pi}{180^0}=\frac{2\pi}{3}=\frac{2}{3}\pi \\ 240^0\times\frac{\pi}{180^0}=\frac{4\pi}{3}=\frac{4}{3}\pi \end{gathered}[/tex]For
[tex]\begin{gathered} 2\sin x+\sqrt[]{3}=0 \\ 2\sin x=-\sqrt[]{3} \\ \Rightarrow\sin x=-\frac{\sqrt[]{3}}{2} \\ x=\sin ^{-1}(-\frac{\sqrt[]{3}}{2}) \\ \therefore x\Rightarrow240^0,300^0 \end{gathered}[/tex]In radian,
[tex]\begin{gathered} 300^0\times\frac{\pi}{180^0}=\frac{5\pi}{3}=\frac{5}{3}\pi \\ \end{gathered}[/tex]Hence, the solutions in radians of the equation in the interval [0, 2π) is
[tex]\frac{2}{3}\pi,\frac{4}{3}\pi\text{ and }\frac{5}{3}\pi[/tex]