Find all solutions of the following equation in the interval [0, 2π).
B)0
C)
D)

Question

Find all solutions of the following equation in the interval [0, 2π). ​  ​
B)0 
C)​ 
D)​

Quizplus · Accepted Answer

We can start by simplifying the equation using the identity $2\sin x\cos x = \sin 2x$. This gives us: $\sin 2x - \dfrac{\sin 4x}{4} = 0$.

Now let $t = \cos 2x$. Then, $\sin 2x = \pm \sqrt{1-t^2}$ and $\sin 4x = 2\sin 2x \cos 2x = \pm 2\sqrt{1-t^2}t$.

Substituting these back into the original equation, we get: $\pm \sqrt{1-t^2} - \dfrac{\pm 2\sqrt{1-t^2}t}{4} = 0$.

Simplifying, we have: $\sqrt{1-t^2}(1-t) = 0$. This has solutions when $\sqrt{1-t^2} = 0$ or $1-t = 0$.

When $\sqrt{1-t^2} = 0$, we have $t = \pm 1$. This corresponds to $2x = 2k\pi$ or $2x = (2k+1)\pi$, where $k$ is an integer. However, we need to check that $0 \leq 2x < 2\pi$ for each of these solutions. We find that $2x = 0$ and $2x = \pi$ work, but $2x = 2\pi$ and $2x = 3\pi$ do not.

When $1-t = 0$, we have $t = 1$, which corresponds to $2x = 2n\pi$ for some integer $n$. Checking $0 \leq 2x < 2\pi$, we find that $2x = 0$ and $2x = 2\pi$ are solutions.

Therefore, the only solutions in $[0, 2\pi)$ are $x = 0$ and $x = \pi/2$, which correspond to choices $\boxed{	extbf{(B)}}$.

Find All Solutions of the Following Equation in the Interval