Find the exact solutions of the given equation in the interval [0, 2$\pi$). $\cos 2 x + 3 \cos x + 2 = 0$
A) $x = 0 , \pi , \frac { 2 \pi } { 3 } , \frac { 5 \pi } { 3 }$
B) $x = \pi , \frac { \pi } { 3 } , \frac { 5 \pi } { 3 }$
C) $x = \pi , \frac { 2 \pi } { 3 } , \frac { 4 \pi } { 3 }$
D) x = 0
E) $x = 0 , \pi , \frac { \pi } { 3 } , \frac { 4 \pi } { 3 }$

Question

Find the exact solutions of the given equation in the interval [0, 2$\pi$).   $\cos 2 x + 3 \cos x + 2 = 0$
A)  $x = 0 , \pi , \frac { 2 \pi } { 3 } , \frac { 5 \pi } { 3 }$
B)  $x = \pi , \frac { \pi } { 3 } , \frac { 5 \pi } { 3 }$
C)  $x = \pi , \frac { 2 \pi } { 3 } , \frac { 4 \pi } { 3 }$
D) x = 0
E)  $x = 0 , \pi , \frac { \pi } { 3 } , \frac { 4 \pi } { 3 }$

Quizplus · Accepted Answer

To solve the equation $\cos 2x + 3\cos x + 2 = 0$, we use the double-angle identity for cosine, which is $\cos 2x = 2\cos^2 x - 1$. Substituting this into the equation gives $2\cos^2 x - 1 + 3\cos x + 2 = 0$, simplifying to $2\cos^2 x + 3\cos x + 1 = 0$. Factoring this quadratic equation in terms of $\cos x$ yields $(2\cos x + 1)(\cos x + 1) = 0$, leading to $\cos x = -\frac{1}{2}$ or $\cos x = -1$. For $\cos x = -\frac{1}{2}$, the solutions in the interval $[0, 2\pi)$ are $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$. For $\cos x = -1$, the solution is $x = \pi$. Therefore, the exact solutions are $x = \pi, \frac{2\pi}{3}, \frac{4\pi}{3}$.

Find the Exact Solutions of the Given Equation in the Interval