a. Sinx-sin2x+sin3x-sin4x=0 b. Sin2x-căn3cos2x=căn3-2cosx

`a) sin x - sin 2x + sin 3x - sin 4x = 0`

`-> (sin x - sin 4x) - (sin 2x - sin 3x) = 0`

`-> 2.cos ((5x)/2).sin (-(3x)/2) - 2.cos ((5x)/2).sin (-x/2) = 0`

`-> 2cos ((5x)/2)[sin ((-3x)/2 - sin (-x/2)] = 0`

`->` \(\left[ \begin{array}{l}2cos \dfrac{5x}{2} = 0\\sin \dfrac{-3x}{2} = sin \dfrac{-x}{2}\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}\dfrac{5x}{2} = \dfrac{\pi}{2} + k\pi\\\dfrac{-3x}{2} = \dfrac{-x}{2} + k2\pi\\\dfrac{-3x}{2} = \pi + \dfrac{x}{2} + k2\pi\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x = \dfrac{\pi}{5} + k\dfrac{2\pi}{5}\\x = k2\pi\\x = \dfrac{-\pi}{2} + k4\pi\end{array} \right.\) `(k in ZZ)`

`b) sin 2x - sqrt{3}cos 2x = sqrt{3} - 2cos x`

`-> 2sin x.cos x + 2cos x - sqrt{3}(1 - 2sin^2 x) - sqrt{3} = 0`

`-> 2cos x(sin x + 1) - sqrt{3}(1 - 2sin^2 x + 1) = 0`

`-> 2[cos x(sin x + 1) + sqrt{3}(sin^2 x - 1) = 0`

`-> cos x(sin x + 1) + sqrt{3}(sin x + 1)(sin x - 1) = 0`

`-> (sin x + 1)(sqrt{3}sin x - sqrt{3} + cos x) = 0`

`->` \(\left[ \begin{array}{l}sin x = -1\\\sqrt{3}sin x + cos x = \sqrt{3}\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x = \dfrac{-\pi}{2} + k\pi\\\dfrac{\sqrt{3}}{2}sin x + \dfrac{1}{2}cos x = \dfrac{\sqrt{3}}{2}\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x = \dfrac{-\pi}{2} + k\pi\\sin (x + \dfrac{\pi}{6}) = \dfrac{\sqrt{3}}{2}\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x = \dfrac{-\pi}{2} + k\pi\\x + \dfrac{\pi}{6} = \dfrac{\pi}{3} + k2\pi\\x + \dfrac{\pi}{6} = \dfrac{2\pi}{3} + k2\pi\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x = x = \dfrac{-\pi}{2} + k\pi\\x = \dfrac{\pi}{6} + k2\pi\\x = \dfrac{\pi}{2} + k2\pi\end{array} \right.\) `(k in ZZ)`