# Thread: Solve a trigonomoetric equation

1. ## Solve a trigonomoetric equation

Hi there - I need to Solve 2sin (2θ) = 1 for 0<x<2pi

I've tried this so far:

2sin2x = 1
sin2x = 1/2
2sinxcosx = 1/2
sinxcosx=1/4
sinxcosx - 1/4 = 0

But I don't know how to factor this, if I should be factoring, or how to get all the possible solutions for x... Could someone give me a hand?

- Cam

2. Originally Posted by Cam5
Hi there - I need to Solve 2sin (2θ) = 1 for 0<x<2pi

I've tried this so far:

2sin2x = 1
sin2x = 1/2
2sinxcosx = 1/2
sinxcosx=1/4
sinxcosx - 1/4 = 0

But I don't know how to factor this, if I should be factoring, or how to get all the possible solutions for x... Could someone give me a hand?

- Cam
$2\sin(2\theta)=1$

$\sin(2\theta)=\frac{1}{2}$

If we restrict the domain of the sine function to:

$\left[-\frac{\pi}{2}\leq 2x \leq \frac{\pi}{2}\right]\rightarrow \left[-\frac{\pi}{4}\leq x \leq \frac{\pi}{4}\right]$,

we can use the inverse sine function to solve for reference angle 2x and then x.

$\sin^{-1}(\sin(2x))=\sin^{-1}\left(\frac{1}{2}\right)$

$2x=\sin^{-1}\left(\frac{1}{2}\right)$

$x=\frac{1}{2} \sin^{-1} \left(\frac{1}{2}\right)$

$x\approx .2617993878$ radians.

3. Hello, Cam5!

You started off fine . . .

Solve: . $2\sin(2\theta) \:=\:1$ .for $0\,<\,x\,<\,2\pi$

I've tried this so far: . $2\sin(2\theta) \:=\: 1 \quad\Rightarrow\quad \sin(2\theta) \:=\: \tfrac{1}{2}$

I'm delighted that you know that double-angle identity,
. . but that's not the way to go . . .

We have: . $\sin(2\theta) \:=\:\tfrac{1}{2}$

. . .Then: . $2\theta \:=\:\frac{\pi}{6},\:\frac{5\pi}{6},\:\frac{13\pi} {6},\:\frac{17\pi}{6},\:\frac{25\pi}{6},\:\frac{29 \pi}{6},\:\hdots$

. . . $\text{Hence: }\;\theta \;=\;\underbrace{\frac{\pi}{12},\:\frac{5\pi}{12}, \:\frac{13\pi}{12},\:\frac{17\pi}{12}}_{\text{Thes e are in the interval}},\:\frac{25\pi}{12},\:\frac{29\pi}{12},\ :\hdots$

Therefore: . $\theta \;=\;\frac{\pi}{12},\;\frac{5\pi}{12},\;\frac{13\p i}{12},\;\frac{17\pi}{12}$