Equazione Goniometrica

$cos(a)cos(b) = \dfrac{1}{2}\cdot cos(a-b) + \dfrac{1}{2}\cdot cos(a+b)$

$sin(a)sin(b) = \dfrac{1}{2}\cdot cos(a-b) - \dfrac{1}{2}\cdot cos(a+b)$

Applicando le formule alla tua equazione:

A sinistra si ottiene

$ \text{cos}\left (\dfrac{4}{3}\pi \right )-\text{cos}\left (2x +\dfrac{4}{3} \pi\right ) $ 

A destra si ottiene

$ \text{cos}\left (\dfrac{5}{3}\pi\right ) + \text{cos} \left (2x +\dfrac{5}{3} \pi \right )  +\sqrt{3} -1$

Quindi

$ \text{cos}\left (\dfrac{4}{3}\pi \right )-\text{cos}\left (2x +\dfrac{4}{3} \pi\right ) = \text{cos}\left (\dfrac{5}{3}\pi\right ) + \text{cos} \left (2x +\dfrac{5}{3} \pi \right )  +\sqrt{3} -1$

Riordina i termini nel seguente modo

$ \left [ cos\left (\dfrac{4}{3}\pi\right ) - cos\left (\dfrac{5}{3}\pi\right ) \right ]  - \left [cos\left (2x +\dfrac{4}{3} \pi\right ) +cos\left (2x +\dfrac{5}{3} \pi \right ) \right]$ =  $+\sqrt{3} -1$

Ora con le formule di prostaferesi tra gli addendi nelle parentesi quadre

$\left [- 2sin\left (\dfrac{3}{2}\pi\right )sin\left (- \dfrac{\pi}{6}\right )\right ] - \left [2cos\left (\dfrac{4x + 3\pi}{2}\right )cos\left (\pi/6\right )\right ]$ = $+\sqrt{3} -1$

$ -\left   [-2\cdot -\dfrac{1}{2} \right ] - 2cos\left (\dfrac{4x + 3\pi}{2} \right )\cdot \dfrac{\sqrt{3}}{2} = +\sqrt{3} -1$

$cos \left (2x + \dfrac{3}{2}\pi \right ) = - \left (\sqrt{3} -1 + 1 \right ) \dfrac{2}{2\cdot \sqrt{3}}$

$sin \left (2x  \right ) = - 1$

Ora rimane da risolvere un'equazione elementare. 

Con le formule di Werner
* sin(a)*sin(b) = (cos(a - b) - cos(a + b))/2
* cos(a)*cos(b) = (cos(a - b) + cos(a + b))/2
si ha
* sin(x)*sin(x + 4*π/3) = (cos(- 4*π/3) - cos(2*x + 4*π/3))/2 = - (1/2 + cos(2*x + 4*π/3))/2
* cos(x)*cos(5*π/3 + x) = (cos(- 5*π/3) + cos(2*x + 5*π/3))/2 = + (1/2 + cos(2*x + 5*π/3))/2
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Con la formula di addizione del coseno
* cos(a + b) = cos(a)*cos(b) - sin(a)*sin(b)
si ha
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* sin(x)*sin(x + 4*π/3) = - (1/2 + cos(2*x + 4*π/3))/2 =
= - (1/2 + cos(2*x)*cos(4*π/3) - sin(2*x)*sin(4*π/3))/2 =
= - (1/2 + cos(2*x)*(- 1/2) - sin(2*x)*(- √3/2))/2 =
= (cos(2*x) - (√3)*sin(2*x) - 1)/4
---------------
* cos(x)*cos(5*π/3 + x) = + (1/2 + cos(2*x + 5*π/3))/2 =
= (1/2 + cos(2*x)*cos(5*π/3) - sin(2*x)*sin(5*π/3))/2 =
= (1/2 + cos(2*x)*(1/2) - sin(2*x)*(- √3/2))/2 =
= (cos(2*x) + (√3)*sin(2*x) + 1)/4
------------------------------
EQUAZIONE
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* sin(x)*sin(x + 4*π/3) = cos(x)*cos(5*π/3 + x) + (√3 - 1)/2 ≡
≡ (cos(2*x) - (√3)*sin(2*x) - 1)/4 = (cos(2*x) + (√3)*sin(2*x) + 1)/4 + (√3 - 1)/2 ≡
≡ (cos(2*x) + (√3)*sin(2*x) + 1)/4 - (cos(2*x) - (√3)*sin(2*x) - 1)/4 + (√3 - 1)/2 = 0 ≡
≡ ((√3)*sin(2*x) + 1)/2 + (√3 - 1)/2 = 0 ≡
≡ (√3)*sin(2*x) + 1 + √3 - 1 = 0 ≡
≡ sin(2*x) = - 1 ≡
≡ 2*x = 2*k*π + arcsin(- 1) ≡
≡ x = k*π + arcsin(- 1)/2 ≡
≡ x = k*π - π/4
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CONTROPROVA
http://www.wolframalpha.com/input?i=sin%28x%29*sin%28x--4*%CF%80%2F3%29%3Dcos%28x%29*cos%285*%CF%80%2F3--x%29%2B%28%E2%88%9A3-1%29%2F2

SIN(x)·SIN(x + 4/3·pi) = COS(x)·COS(5/3·pi + x) + (√3 - 1)/2

SIN(x + 4/3·pi) = - SIN(x + pi/3)

COS(5/3·pi + x) = SIN(x + pi/6)

COS(x)·SIN(x + pi/6) + SIN(x)·SIN(x + pi/3) + (√3 - 1)/2 = 0

SIN(x + pi/6) = SIN(x)·COS(pi/6) + SIN(pi/6)·COS(x)=

=COS(x)/2 + √3·SIN(x)/2

SIN(x + pi/3) = SIN(x)·COS(pi/3) + SIN(pi/3)·COS(x)=

=√3·COS(x)/2 + SIN(x)/2

Quindi:

COS(x)·(COS(x)/2 + √3·SIN(x)/2) + SIN(x)·(√3·COS(x)/2 + SIN(x)/2) + (√3 - 1)/2 = 0

Poniamo:

Χ = COS(x)

Υ = SIN(x)

con: Χ^2 + Υ^2 = 1

quindi:

Χ·(Χ/2 + √3·Υ/2) + Υ·(√3·Χ/2 + Υ/2) + (√3 - 1)/2 = 0

Υ^2/2 + √3·Υ·Χ + Χ^2/2 + √3/2 - 1/2 = 0

Υ^2 + 2·√3·Υ·Χ + Χ^2 + √3 - 1 = 0

2·√3·Υ·Χ + √3 = 0

2·Υ·Χ + 1 = 0

2·SIN(x)·COS(x) + 1 = 0

SIN(2·x) = -1

2·x = 3/2·pi + 2·k·pi----->x = k·pi + 3·pi/4