Question

How do I solve this equation:
sin( (2π/3)-x ) = sin( x )
(0<=x<=7π/3)

Answer 1

\[\sf \begin{align*}
\sin\left(\frac{2\pi}{3} -x\right)=\sin(x)\\
&:~ \color{red}{\sin(\alpha-\beta) = \sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)}\\
&: ~ \color{blue}{\sin\left(\frac{2\pi}{3}\right)}\cos(x)-\color{blue}{\cos\left(\frac{2\pi}{3}\right)}\sin(x)=\sin(x)\\
&:\frac{\sqrt{3}}{2}\cos(x)+\left(\frac{1}{2}\right)\sin(x)=\sin(x)\\
&: ~ \frac{\sqrt{3}}{2}\cos(x) = \frac{1}{2}\sin(x)\\
\end{align*}\]

Answer 2

Ah... hmm... We want constants on one side, but our function on the other. We can divide both sides by either cosine or sine.

I'll divide both sides by cos(x)

Answer 3

\[\begin{align*}
\frac{\sqrt{3}}{2}\cos(x)=\frac{1}{2}\sin(x)
\\&: ~ \sqrt{3}=\tan(x)
\\&: ~ \tan^{-1}(\sqrt{3} ) = \tan^{-1}(\tan(x))
\\&: ~ \tan^{-1}(\sqrt{3}) = x
\\&: ~ x=\frac{\pi}{3}
\end{align*}\]Since we are trying to find all the values between \(0\) and \(\dfrac{7\pi}{3}\) We need to find the general solution for x, For tangent functions, we know the period is \(\pi\) so starting at \(\dfrac{\pi}{3}\) and adding it to our period, \(k\pi\) where \(k=1,2,3...\) We can state that: \(\boxed{x=\dfrac{\pi}{3} +k\pi} \) is our solution.

Answer 4

hmm...
try expanding LHS:

sin (2pi/3) cos x - cos (pi/3) sinx = sin x

Answer 5

sin 2pi/3 = sqrt3/2 and cos 2pi/3 = -1/2

Answer 6

giving us
(qrt3/2) cos x - (1/2) sin x = 0

Answer 7

(sqrt3 / 2) cos x = (1/2) sin x
divide both sides by cos x
sqrt3/2 = 1/2 tan x

solve for x

Answer 8

Another way:
From
\(\large \sin\left(\frac{2\pi}{3} -x\right)=\sin(x)\\\)

we deduce
\(\large \left(\frac{2\pi}{3} -x\right)=x+2k\pi\\\) where \(k\in Z\)
Solve for x
\(\large x=\frac{\pi}{3}+k\pi\)
as before, but limit values of k for \([0\le x \le 7\pi/3]\)