Question

3cosx-4sinx=2

cosx-sinx=1

3sin2x+4cos2x=5

Answer 1

Решается введением вспомогательного угла.

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

 

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

 

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