Find all solutions of $2 \sin x - \sqrt { 2 } = 0$
A) $x = \frac { \pi } { 4 }$ and $x = \frac { 3 \pi } { 4 }$

B) $x = \frac { \pi } { 6 } + 2 k \pi$ and $x = \frac { \pi } { 2 } + 2 k \pi , k \in \text { integer }$

C) $x = \frac { \pi } { 3 } + 2 k \pi$ and $x = \frac { \pi } { 2 } + 2 k \pi , k \in \text { integer }$

D) $x = \frac { \pi } { 2 } \text { and } x = \frac { 3 \pi } { 2 }$

E) $x = \frac { \pi } { 4 } + 2 k \pi$ and $x = \frac { 3 \pi } { 4 } + 2 k \pi , k \in \text { integer }$

Question

Find all solutions of $2 \sin x - \sqrt { 2 } = 0$
A) $x = \frac { \pi } { 4 }$ and $x = \frac { 3 \pi } { 4 }$

B) $x = \frac { \pi } { 6 } + 2 k \pi$ and $x = \frac { \pi } { 2 } + 2 k \pi , k \in \text { integer }$

C) $x = \frac { \pi } { 3 } + 2 k \pi$ and $x = \frac { \pi } { 2 } + 2 k \pi , k \in \text { integer }$

D) $x = \frac { \pi } { 2 } \text { and } x = \frac { 3 \pi } { 2 }$

E) $x = \frac { \pi } { 4 } + 2 k \pi$ and $x = \frac { 3 \pi } { 4 } + 2 k \pi , k \in \text { integer }$

Quizplus · Accepted Answer

Solving $2 \sin x - \sqrt { 2 } = 0$ gives $\sin x = \frac{\sqrt{2}}{2}$, which corresponds to $x = \frac{\pi}{4}$ and $x = \frac{3\pi}{4}$ in the principal interval $[0, 2\pi)$. Including the periodic nature of the sine function, the general solutions are $x = \frac{\pi}{4} + 2k\pi$ and $x = \frac{3\pi}{4} + 2k\pi$, where $k$ is an integer.

Find All Solutions Of 2sin⁡x−2=02 \sin x - \sqrt { 2 } = 02sinx−2​=0

Find All Solutions Of $2 \sin x - \sqrt { 2 } = 0$