Ответы и объяснения

2013-05-21T01:58:08+04:00

Решается введением вспомогательного угла.

3cosx-4sinx=2\\R=\sqrt{3^2+4^2}=\sqrt{25}=5\\\frac{3}{5}cosx-\frac{4}{5}sinx=\frac{2}{5}\\sin\beta=\frac{3}{5}\ \ \ \ \ cos\beta=\frac{4}{5}\\sin\beta cosx-cos\beta sinx=\frac{2}{5}\\sin(\beta-x)=\frac{2}{5}\\\beta-x=(-1)^n*arcsin(\frac{4}{5})+\pi*n\\-x=(-1)^n*arcsin(\frac{2}{5})+\pi*n-\beta\\x=(-1)^{n+1}*arcsin(\frac{2}{5})-\pi*n+\arcsin(\frac{3}{5})

 

cosx-sinx=1\\R=\sqrt{1^2+1^2}=\sqrt{2}\\\frac{1}{\sqrt{2}}cosx-\frac{1}{\sqrt{2}}sinx=\frac{1}{\sqrt{2}}\\\frac{\sqrt{2}}{2}cosx-\frac{\sqrt{2}}{2}sinx=\frac{\sqrt{2}}{2}\\sin\frac{\pi}{4}*cosx-\cos\frac{\pi}{4}*sinx=\frac{\sqrt{2}}{2}\\sin(\frac{\pi}{4}-x)=\frac{\sqrt{2}}{2}\\\frac{\pi}{4}-x=(-1)^n*arcsin(\frac{\sqrt{2}}{2})+\pi*n\\x=(-1)^{n+1}*\frac{\pi}{4}-\pi*n+\frac{\pi}{4}

 

3sin2x+4cos2x=5\\R=\sqrt{3^2+4^2}=\sqrt{25}=5\\\frac{3}{5}sin2x+\frac{4}{5}cos2x=1\\cos\beta=\frac{3}{5}\ \ \ \ \ sin\beta=\frac{4}{5}\\cos\beta sin2x+sin\beta cos2x=1\\sin(\beta+2x)=1\\\beta+2x=\pi*n\\2x=\pi*n-\beta\\x=\frac{\pi n}{2}-\frac{arcsin\frac{4}{5}}{2}