91Ó°ÊÓ

We are given the equation \(2 \cos^2 x - 3 \cos x + 1 = 0\). Let's denote \(\cos x = y\). Then the equation becomes a quadratic equation in y: \(2y^2 - 3y + 1 = 0\).

Now, we have to solve the quadratic equation \(2y^2 - 3y + 1 = 0\). The quadratic formula to find the roots is given by \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -3\), and \(c = 1\). Applying the formula, we get \(y = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(1)}}{2(2)} = \frac{3 \pm \sqrt{1}}{4}\). So the roots of the quadratic equation are \(y = \frac{3 - 1}{4} = \frac{1}{2}\), and \(y = \frac{3 + 1}{4} = 1\).

We have found the roots in terms of y (which is equivalent to \(\cos x\)): \(y = \frac{1}{2}\) and \(y = 1\). Now, we will find the values of x for each root within the interval \([0, 2\pi)\). 1. For \(y = \cos x = \frac{1}{2}\): We must find x such that \(\cos x = \frac{1}{2}\). We know that the cosine function has a value of \(\frac{1}{2}\) at angles \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\), both of which are in the given interval. Hence, the solutions for this case are: \(x = \frac{\pi}{3}, \frac{5\pi}{3}\). 2. For \(y = \cos x = 1\): We must find x such that \(\cos x = 1\). We know that the cosine function has a value of 1 at angle \(x = 0\), which is in the given interval. Hence, the solution for this case is: \(x = 0\).

Now we have found all the solutions of the equation \(2\cos^2x - 3\cos x +1 = 0\) within the interval \([0, 2\pi)\). The solutions are \(x = 0, \frac{\pi}{3}, \frac{5\pi}{3}\).