Question

4sin²
`(x)/(2)`=3

Answer 1

`\displaystyle x=\left \{(2\pi)/(3)+2\pi k,(4\pi)/(3)+2\pi k, (8\pi)/(3)+2\pi k, (10\pi)/(3)+2\pi k\right \}k\in \mathbb{Z}`

Explanation:

Hi there!

We want to solve for
`x` in:

`4\sin^2((x)/(2))=3`

Since
`x` is in the argument of
`\sin^2`, let's first isolate
`\sin^2` by dividing both sides by 4:

`\displaystyle \sin^2\left((x)/(2)\right)=(3)/(4)`

Next, recall that
`\sin^2x` is just shorthand notation for
`(\sin x)^2`. Therefore, take the square root of both sides:

`\displaystyle \sqrt{\sin^2\left((x)/(2)\right)}=\sqrt{(3)/(4)},\\\sin\left((x)/(2)\right)=\pm \sqrt{(3)/(4)}`

Simplify using
`\displaystyle \sqrt{(a)/(b)}=(√(a))/(√(b))`:

`\displaystyle \sin\left((x)/(2)\right)=\pm \sqrt{(3)/(4)},\\\sin\left((x)/(2)\right)=\pm (√(3))/(√(4))=\pm (√(3))/(2)`

Let
`\phi = (x)/(2)`.

Case 1 (positive root):

`\displaystyle \sin(\phi)=(√(3))/(2),\\\phi = (\pi)/(3)+2\pi k, k\in \mathbb{Z}, \\\\\phi =(2\pi)/(3)+2\pi k, k\in \mathbb{Z}`

Therefore, we have:

`\displaystyle (x)/(2)=\phi = (\pi)/(3)+2\pi k, k\in \mathbb{Z}, \\\\(x)/(2)=\phi =(2\pi)/(3)+2\pi k, k\in \mathbb{Z},\\\\\begin{cases}x=\boxed{(2\pi)/(3)+2\pi k, k\in \mathbb{Z}},\\x=\boxed{(4\pi)/(3)+2\pi k , k \in \mathbb{Z}}\end{cases}`

Case 2 (negative root):

`\displaystyle \sin(\phi)=-(√(3))/(2),\\\phi = (4\pi)/(3)+2\pi k, k\in \mathbb{Z}, \\\\\phi =(5\pi)/(3)+2\pi k, k\in \mathbb{Z},\\\begin{cases}x=\boxed{(8\pi)/(3)+2\pi k, k\in \mathbb{Z}},\\x=\boxed{(10\pi)/(3)+2\pi k , k \in \mathbb{Z}}\end{cases}`