solve 5 sin(π/4 x)=3 for the four smallest positive solutions. x = give your answers accurate to at least…

Explanation:

Step1: Isolate the sine - function

Divide both sides of the equation $5\sin(\frac{\pi}{4}x)=3$ by 5. We get $\sin(\frac{\pi}{4}x)=\frac{3}{5}=0.6$.

Step2: Find the principal inverse - sine value

Let $\theta = \frac{\pi}{4}x$. Then $\theta=\sin^{-1}(0.6)$. Using a calculator, $\sin^{-1}(0.6)\approx0.6435$ radians.

Step3: Use the unit - circle properties for sine function

The sine function $y = \sin\theta$ has a period of $2\pi$. The general solution for $\sin\theta = 0.6$ is $\theta=\sin^{-1}(0.6)+2k\pi$ or $\theta=\pi-\sin^{-1}(0.6)+2k\pi$, where $k\in\mathbb{Z}$. For $\theta=\frac{\pi}{4}x$:

Case 1: $\frac{\pi}{4}x=\sin^{-1}(0.6)+2k\pi$

$x=\frac{4}{\pi}(\sin^{-1}(0.6)+2k\pi)$.

Case 2: $\frac{\pi}{4}x=\pi - \sin^{-1}(0.6)+2k\pi$

$x=\frac{4}{\pi}(\pi-\sin^{-1}(0.6)+2k\pi)$.

Step4: Find the four smallest positive solutions

When $k = 0$ in $\frac{\pi}{4}x=\sin^{-1}(0.6)$: $x_1=\frac{4}{\pi}\sin^{-1}(0.6)\approx\frac{4}{\pi}\times0.6435\approx0.82$. When $k = 0$ in $\frac{\pi}{4}x=\pi-\sin^{-1}(0.6)$: $x_2=\frac{4}{\pi}(\pi - \sin^{-1}(0.6))=\frac{4}{\pi}(\pi - 0.6435)\approx3.18$. When $k = 1$ in $\frac{\pi}{4}x=\sin^{-1}(0.6)+2\pi$: $x_3=\frac{4}{\pi}(\sin^{-1}(0.6)+2\pi)=\frac{4}{\pi}(0.6435 + 2\pi)\approx8.82$. When $k = 1$ in $\frac{\pi}{4}x=\pi-\sin^{-1}(0.6)+2\pi$: $x_4=\frac{4}{\pi}(\pi-\sin^{-1}(0.6)+2\pi)=\frac{4}{\pi}(\pi - 0.6435+2\pi)\approx11.18$.

Answer:

$0.82,3.18,8.82,11.18$