a) Przekształćmy równanie

$\sin 2x+\sqrt{3}\cos x=0$  

$2\sin x\cos x+\sqrt{3}\cos x=0$  

$\cos x\left(2\sin x+\sqrt{3}\right)=0$  

$\cos x=0\text{lub}2\sin x+\sqrt{3}=0$  

$1.\cos x=0\text{lub}2.\sin x=-\frac{\sqrt{3}}{2}$

 

$1.x=\frac{\pi }{2}+k\pi ,k\in \mathbb{Z}$   

 

$2.\sin x=-\frac{\sqrt{3}}{2}$  

$-\sin x=\frac{\sqrt{3}}{2}$  

$\sin \left(\pi +x\right)=\frac{\sqrt{3}}{2}$  

Stąd

$\pi +x=\frac{\pi }{3}+2k\pi \text{lub}\pi +x=\frac{2\pi }{3}+2k\pi ,k\in \mathbb{Z}$  

$x=-\frac{2\pi }{3}+2k\pi \text{lub}x=-\frac{\pi }{3}+2k\pi ,k\in \mathbb{Z}$  

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b) Mamy

$\frac{\cos x}{\sin 2x}=-\frac{\sqrt{2}}{2}$  

$\frac{\cos x}{2\sin x\cos x}=-\frac{\sqrt{2}}{2}$  

$\frac{\cos x}{\sin x\cos x}=-\sqrt{2}$  

$\frac{1}{\sin x}=-\sqrt{2}$  

$-\frac{1}{\sqrt{2}}=\sin x$  

$\sin x=-\frac{\sqrt{2}}{2}$  

$-\sin x=\frac{\sqrt{2}}{2}$  

Skorzystamy ze wzoru redunkcyjnego

$\sin \left(\pi +x\right)=\frac{\sqrt{2}}{2}$  

Zatem

$\pi +x=\frac{\pi }{4}+2k\pi \text{lub}\pi +x=\frac{3\pi }{4}+2k\pi ,k\in \mathbb{Z}$

  $x=-\frac{3\pi }{4}+2k\pi \text{lub}x=-\frac{\pi }{4}+2k\pi ,k\in \mathbb{Z}$

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c) Mamy

$\sin 2x=2{\sin }^{2}x$  

$\sin 2x-2{\sin }^{2}x=0$  

$2\sin x\cos x-2{\sin }^{2}x=0$  

$2\sin x\left(\cos x-\sin x\right)=0$  

$2\sin x=0\text{lub}\cos x-\sin x=0$  

$1.\sin x=0\text{lub}2.\cos x=\sin x$  

 

$1.x=k\pi ,k\in \mathbb{Z}$

 

Dla równania 2. narysujemy rysunek.

$2.x=\frac{\pi }{4}+k\pi ,k\in \mathbb{Z}$  

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d) Mamy

$\sin 2x=2\cos x$  

$\sin 2x-2\cos x=0$  

$2\sin x\cos x-2\cos x=0$  

$2\cos x\left(\sin x-1\right)=0$  

$2\cos x=0\text{lub}\sin x-1=0$  

$\cos x=0\text{lub}\sin x=1$  

Stąd

$x=\frac{\pi }{2}+k\pi \text{lub}x=\frac{\pi }{2}+2k\pi ,k\in \mathbb{Z}$  

$x=\frac{\pi }{2}+k\pi ,k\in \mathbb{Z}$