Solve the following trigonometric equation for \( x \): \[2\sin(x)\cos(x)+\sin(x)-1=0\]

Step 1 ：First, we rewrite the equation using the double-angle identity \(\sin(2x)=2\sin(x)\cos(x)\): \[\sin(2x) + \sin(x) - 1 = 0\]

Step 2 ：Next, we rearrange the equation in a standard quadratic form: \[\sin(2x) + \sin(x) - 1 = 0 \Rightarrow \sin(x)^2 + \sin(x) - 1 = 0\]

Step 3 ：Now, we can apply the quadratic formula to solve for \(\sin(x)\): \[\sin(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}\]

Step 4 ：Finally, we solve for \(x\) by taking the inverse sine of each solution: \[x = \sin^{-1}\left(\frac{-1 + \sqrt{5}}{2}\right), \sin^{-1}\left(\frac{-1 - \sqrt{5}}{2}\right)\]