# Trig identities

## Homework Statement

2sin(2x) - $$\sqrt{3}$$ = 0

Find the value of the variable.

## Homework Equations

Sin(2x) = 2SinxCosx

## The Attempt at a Solution

I subtracted root3 and then divided everything by 2 which leaves me with

sin (2x) = $$\sqrt{3}$$/2

then I used the double angle identity to change sin 2x into 2sinxcosx.

I'm stuck from there. I forgot how to use double angles ( eek.)

jgens
Gold Member
I wouldn't use the double angle identity to solve this problem. Go back to, sin(2x) = (3)1/2/2 and take the inverse sine.

How do I do that?

jgens
Gold Member
If asked you to find all x that satisfy, sin(x) = 1/2, what would you do? If you can do that problem, then you can do this one.

I'm still stuck. I used the inverse sin of the (root3)/2 and I got answers like pi/3, 2pi/3, ect.

I think I need to use the identity but it seems impossible to figure out:

2sinx cosx = (root3)/2

jgens
Gold Member
You don't need the identity. You got multiple answers because the trigonmetic functions are periodic so there are an infinite number of solutions to equations like sin(x) = 1/2. You really want to use the inverse sin in this case.

What happens to the 2x ?

jgens
Gold Member
You end up solving it for x, just the same as if I told you 2x = 6.

HOLY oh man, I am real angry right now.

I left my calculator in radian mode and I was using degrees for like the last 40 mins !

Whats significance does the identity: sin2x = 2sinxcosx have?

jgens
Gold Member
Whats significance does the identity: sin2x = 2sinxcosx have?

It doesn't have significance in this problem. It's useful when you need to manipulate trigonometric functions to put them into an integrable form or when you need to prove other identities.

let t=2x.

can you solve sin(t) = sqrt(3)/2

yeah, its pi/6, 5pi/6, and so forth.
Whats the sign that I add to the answer if it's periodical? How exactly do I write the answer for this problem? Do I write all the radians that make the equation true one time around the unit circle then add the (periodical)?

You usually write your answers with k where k is any integer; for example, solutions for sin(x) = 1 is x = pi/2 + 2k*pi

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You should end up with multiple answers as said before trig functions are periodical.
Did the question have any limits? i.e 0<x<2pi/360??

No limits.

So my answer is basicaly, 30 degrees or pi/6 + 2kpi? Correct?

pi/6 + k*pi takes care of half of the answers; you also need the solutions with 5pi/6.

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$$sin(2x)=\frac{\sqrt{3}}{2}$$

$$2x=\begin{bmatrix} \frac{\pi}{3} + 2k\pi\\ \pi-\frac{\pi}{3}+2k\pi \end{bmatrix}$$

Can you solve it now?

x = pi/6 + 2kpi, 5pi/6 = 2kpi?

On a test or something, would I just leave the answers as so? ( if their correct).

Mark44
Mentor
No.
Starting with Дьявол's work (which I haven't checked, but believe is correct), you have
2x = pi/3 + 2k*pi, or 2x = 2pi/3 + 2k*pi, so
x = pi/6 + k*pi, or x = pi/3 + k*pi, where k is any integer.

You forgot to divide the 2k*pi parts.

Mark44 you're right. The final answers are:
x = pi/6 + k*pi, or x = pi/3 + k*pi, where k is any integer.

He forgot to divede the 2k*pi parts.

Regards.

Yups

hullo..??...
sin(2x)=sqrt(3/4)..u can get a genenarl solution 4 dat..the identity is unnecessary over here..
sin t=sin(a)
then t=n(pi)+((-1)^n)a
use this and solve 4 "x"

HallsofIvy