Answer:
[tex]\displaystyle x=\left \{\frac{2\pi}{3}+2\pi k,\frac{4\pi}{3}+2\pi k, \frac{8\pi}{3}+2\pi k, \frac{10\pi}{3}+2\pi k\right \}k\in \mathbb{Z}[/tex]
Step-by-step explanation:
Hi there!
We want to solve for [tex]x[/tex] in:
[tex]4\sin^2(\frac{x}{2})=3[/tex]
Since [tex]x[/tex] is in the argument of [tex]\sin^2[/tex], let's first isolate [tex]\sin^2[/tex] by dividing both sides by 4:
[tex]\displaystyle \sin^2\left(\frac{x}{2}\right)=\frac{3}{4}[/tex]
Next, recall that [tex]\sin^2x[/tex] is just shorthand notation for [tex](\sin x)^2[/tex]. Therefore, take the square root of both sides:
[tex]\displaystyle \sqrt{\sin^2\left(\frac{x}{2}\right)}=\sqrt{\frac{3}{4}},\\\sin\left(\frac{x}{2}\right)=\pm \sqrt{\frac{3}{4}}[/tex]
Simplify using [tex]\displaystyle \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}[/tex]:
[tex]\displaystyle \sin\left(\frac{x}{2}\right)=\pm \sqrt{\frac{3}{4}},\\\sin\left(\frac{x}{2}\right)=\pm \frac{\sqrt{3}}{\sqrt{4}}=\pm \frac{\sqrt{3}}{2}[/tex]
Let [tex]\phi = \frac{x}{2}[/tex].
[tex]\displaystyle \sin(\phi)=\frac{\sqrt{3}}{2},\\\phi = \frac{\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\phi =\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z}[/tex]
Therefore, we have:
[tex]\displaystyle \frac{x}{2}=\phi = \frac{\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\frac{x}{2}=\phi =\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z},\\\\\begin{cases}x=\boxed{\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z}},\\x=\boxed{\frac{4\pi}{3}+2\pi k , k \in \mathbb{Z}}\end{cases}[/tex]
[tex]\displaystyle \sin(\phi)=-\frac{\sqrt{3}}{2},\\\phi = \frac{4\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\phi =\frac{5\pi}{3}+2\pi k, k\in \mathbb{Z},\\\begin{cases}x=\boxed{\frac{8\pi}{3}+2\pi k, k\in \mathbb{Z}},\\x=\boxed{\frac{10\pi}{3}+2\pi k , k \in \mathbb{Z}}\end{cases}[/tex]
