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

Question 97

Multiple Choice

Solve the equation. Give a general formula for all the solutions.
- 2cosθ+1=02 \cos \theta + 1 = 0


A) θ=π2+2kπ,θ=3π2+2kπ\theta = \frac { \pi } { 2 } + 2 k \pi , \theta = \frac { 3 \pi } { 2 } + 2 k \pi
B) θ=2π3+2kπ,θ=4π3+2kπ\theta = \frac { 2 \pi } { 3 } + 2 \mathrm { k } \pi , \theta = \frac { 4 \pi } { 3 } + 2 \mathrm { k } \pi
C) θ=2π3+kπ,θ=4π3+kπ\theta = \frac { 2 \pi } { 3 } + \mathrm { k } \pi , \theta = \frac { 4 \pi } { 3 } + \mathrm { k } \pi
D) θ=3π2+kπ\theta = \frac { 3 \pi } { 2 } + \mathrm { k } \pi

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