Solved

Solve the Following Equation 2sinx1=02 \sin x - 1 = 0

Question 206

Multiple Choice

Solve the following equation.
2sinx1=02 \sin x - 1 = 0


A) x=2π3+2nπ and x=4π3+2nπx = \frac { 2 \pi } { 3 } + 2 n \pi \text { and } x = \frac { 4 \pi } { 3 } + 2 n \pi , where n is an integer
B) x=π6+2nπ and x=5π6+2nπx = \frac { \pi } { 6 } + 2 n \pi \text { and } x = \frac { 5 \pi } { 6 } + 2 n \pi , where n is an integer
C) x=π3+2nπ and x=5π3+2nπx = \frac { \pi } { 3 } + 2 n \pi \text { and } x = \frac { 5 \pi } { 3 } + 2 n \pi , where n is an integer
D) x=π4+2nπ and x=5π4+2nπx = \frac { \pi } { 4 } + 2 n \pi \text { and } x = \frac { 5 \pi } { 4 } + 2 n \pi , where n is an integer
E) x=2π3+2nπ and x=4π3+2nπx = \frac { 2 \pi } { 3 } + 2 n \pi \text { and } x = \frac { 4 \pi } { 3 } + 2 n \pi , where n is an integer.

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions