If (2cos x - 1)(3 + 2cos x) = 0 for 0 le x le | vedclass

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(B) Given the equation $(2\cos x - 1)(3 + 2\cos x) = 0$. This implies either $2\cos x - 1 = 0$ or $3 + 2\cos x = 0$. Case $1$: $2\cos x - 1 = 0 \Right

Vedclass · Accepted Answer

$\frac{\pi}{3}, \frac{5\pi}{3}$

Vedclass · Answer

(B) Given the equation $(2\cos x - 1)(3 + 2\cos x) = 0$. This implies either $2\cos x - 1 = 0$ or $3 + 2\cos x = 0$. Case $1$: $2\cos x - 1 = 0 \Rightarrow \cos x = \frac{1}{2}$. For $0 \le x \le 2\pi$,the solutions are $x = \frac{\pi}{3}$ and $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$. Case $2$: $3 + 2\cos x = 0 \Rightarrow \cos x = -\frac{3}{2}$. Since the range of $\cos x$ is $[-1, 1]$,the equation $\cos x = -\frac{3}{2}$ has no real solution. Thus,the values of $x$ in the interval $[0, 2\pi]$ are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$.

If $(2\cos x - 1)(3 + 2\cos x) = 0$ for $0 \le x \le 2\pi$,then $x = $

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