question 7 of 10\nhow many solutions does the equation sin(4x) = 1/2 have on the interval (0, 2π?\nanswer here

Answer:

8

Explanation:

Step1: Recall sine - value solutions

We know that if $\sin\theta=\frac{1}{2}$, then $\theta = \frac{\pi}{6}+ 2k\pi$ or $\theta=\frac{5\pi}{6}+2k\pi$, where $k\in\mathbb{Z}$.

Step2: Substitute $\theta = 4x$

For $\sin(4x)=\frac{1}{2}$, we have $4x=\frac{\pi}{6}+2k\pi$ or $4x=\frac{5\pi}{6}+2k\pi$.

Step3: Solve for $x$ in the first case

From $4x=\frac{\pi}{6}+2k\pi$, we get $x=\frac{\pi}{24}+\frac{k\pi}{2}$.

Step4: Solve for $x$ in the second case

From $4x=\frac{5\pi}{6}+2k\pi$, we get $x = \frac{5\pi}{24}+\frac{k\pi}{2}$.

Step5: Find valid $k$ values for the interval $(0,2\pi]$

For $x=\frac{\pi}{24}+\frac{k\pi}{2}>0$ and $x=\frac{\pi}{24}+\frac{k\pi}{2}\leq2\pi$. Solving $\frac{\pi}{24}+\frac{k\pi}{2}\leq2\pi$ gives $\frac{1 + 12k}{24}\leq2$, so $1+12k\leq48$, $12k\leq47$, $k\leq\frac{47}{12}\approx3.92$. When $k = 0,1,2,3$, we get valid $x$ values. For $x=\frac{5\pi}{24}+\frac{k\pi}{2}>0$ and $x=\frac{5\pi}{24}+\frac{k\pi}{2}\leq2\pi$. Solving $\frac{5\pi}{24}+\frac{k\pi}{2}\leq2\pi$ gives $\frac{5 + 12k}{24}\leq2$, so $5+12k\leq48$, $12k\leq43$, $k\leq\frac{43}{12}\approx3.58$. When $k = 0,1,2,3$, we get valid $x$ values. The total number of solutions is $4 + 4=8$.