Question

Cos3x-sinx=корень из 3 (cosx-sin3x)

Answer 1

$\cos3x-\sin x= \sqrt{3} (\cos x-\sin 3x)\\ \cos3x-\sin x=\sqrt{3} \cos x-\sqrt{3} \sin3x\\ \cos3x+\sqrt{3} \sin3x=\sin x+\sqrt{3} \cos x\\ \sqrt{(\sqrt{3} )^2+1^2} \sin(3x+\arcsin \frac{1}{ \sqrt{(\sqrt{3} )^2+1^2} } )= \sqrt{(\sqrt{3} )^2+1^2} \sin(x+ \arcsin\frac{\sqrt{3}}{\sqrt{(\sqrt{3})^2+1^2}}$

$2\sin(3x+ \frac{\pi}{6} )=2\sin(x+\frac{\pi}{3}) \\ \\ \sin(3x+\frac{\pi}{6} )=\sin(x+\frac{\pi}{3}) \\ \\ \sin(3x+\frac{\pi}{6} )-\sin(x+\frac{\pi}{3})=0\\ \\ 2\cos \frac{3x+\frac{\pi}{6} +x+\frac{\pi}{3} }{2} \sin \frac{3x+\frac{\pi}{6} -x-\frac{\pi}{3} }{2} =0\\ \\ 2\cos \frac{4x+\frac{\pi}{2} }{2} \sin \frac{2x-\frac{\pi}{6} }{2} =0\\ \\ 2\cos(2x+\frac{\pi}{4} )\sin(x-\frac{\pi}{12} )=0$

Произведение равно нулю, если один из множителей равен нулю

$\cos(2x+\frac{\pi}{4} )=0\\ \\ 2x+\frac{\pi}{4}=\frac{\pi}{2}+ \pi n,n \in Z\\ \\ 2x=\frac{\pi}{2}-\frac{\pi}{4}+ \pi n,n \in Z \\ \\ 2x=\frac{\pi}{4}+ \pi n,n \in Z\\ \\ \boxed{x_1= \frac{\pi}{8} +\frac{\pi n}{2},n \in Z}$

$\sin(x-\frac{\pi}{12})=0\\ \\ x-\frac{\pi}{12}= \pi k,k \in Z\\ \\ \boxed{x_2=\frac{\pi}{12}+ \pi k,k \in Z}$