Solve the equation on the interval [0, 2π).
-Suppose $f ( x ) = 2 \cos \theta + 1$. Solve $f ( x ) = 0$.
A) $\frac { \pi } { 3 } , \frac { 5 \pi } { 3 }$
B) $\frac { \pi } { 2 } , \frac { 3 \pi } { 2 }$
C) $\frac { 3 \pi } { 2 }$
D) $\frac { 2 \pi } { 3 } , \frac { 4 \pi } { 3 }$

Question

Quizplus · Accepted Answer

To solve $2 \cos \theta + 1 = 0$, we first isolate $\cos \theta$ to get $\cos \theta = -\frac{1}{2}$. The values of $\theta$ that satisfy this equation in the interval $[0, 2\pi)$ are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$, which correspond to the angles where the cosine function equals $-\frac{1}{2}$ within one full rotation of the unit circle.

Solve the Equation on the Interval [0, 2π) f(x)=2cos⁡θ+1f ( x ) = 2 \cos \theta + 1f(x)=2cosθ+1

Solve the Equation on the Interval [0, 2π) $f ( x ) = 2 \cos \theta + 1$