Solve the equation. Give a general formula for all the solutions.
-$2 \cos \theta + 1 = 0$
A) $\theta = \frac { \pi } { 2 } + 2 \mathrm { k } \pi , \theta = \frac { 3 \pi } { 2 } + 2 \mathrm { k } \pi$
B) $\theta = \frac { 2 \pi } { 3 } + 2 \mathrm { k } \pi , \theta = \frac { 4 \pi } { 3 } + 2 \mathrm { k } \pi$
C) $\theta = \frac { 2 \pi } { 3 } + \mathrm { k } \pi , \theta = \frac { 4 \pi } { 3 } + \mathrm { k } \pi$
D) $\theta = \frac { 3 \pi } { 2 } + \mathrm { k } \pi$

Question

Quizplus · Accepted Answer

First, solve the equation $2 \cos \theta + 1 = 0$ for $\cos \theta$, which gives $\cos \theta = -\frac{1}{2}$. The general solutions for $\cos \theta = -\frac{1}{2}$ are $\theta = \frac{2\pi}{3} + 2k\pi$ and $\theta = \frac{4\pi}{3} + 2k\pi$, where $k$ is any integer. This corresponds to the angles in the second and third quadrants where the cosine value is $-\frac{1}{2}$.

Solve the Equation 2cos⁡θ+1=02 \cos \theta + 1 = 02cosθ+1=0 A) θ=π2+2kπ,θ=3π2+2kπ\theta = \frac { \pi } { 2 } + 2 \mathrm { k } \pi , \theta = \frac { 3 \pi } { 2 } + 2 \mathrm { k } \piθ=2π​+2kπ,θ=23π​+2kπ

Solve the Equation $2 \cos \theta + 1 = 0$ A) $\theta = \frac { \pi } { 2 } + 2 \mathrm { k } \pi , \theta = \frac { 3 \pi } { 2 } + 2 \mathrm { k } \pi$