Question

cos 5x+ cos 2x+ cos3x + cos4x = 0

Answer 1

$(cos5x+cos3x)+(cos2x+cos4x)=0\ 2cos4x*cosx+2cos3x*cosx=0\ cosx(2cos4x+2cos3x)=0\ left { {{cosx=0} atop {cos4x+cos3x=0}} right. \ \ left { {{x_{1}=pi *n-frac{pi}{2}} atop {cos^4x+sin^4x-6sin^2xcos^2x+cos^3x-3sin^2xcosx=0}} right. \ \ cos^4x+(1-cos^2x)^2-6(1-cos^2x)cos^2x+cos^3x-3(1-cos^2x)cosx=0\ cosx=t\ 8t^4+4t^3-8t^2-3t+1=0\ (t+1)(8t^3-4t^2-4t+1)=0\ t=-1\ cosx=-1\ x_{2}=2pi n+pi$
$cos4x+cos3x=0\ 2cosfrac{7x}{2}*cosfrac{x}{2}=0\ cosfrac{7x}{2}=0\ x=-frac{pi}{7}-frac{2pi*n}{7}\ cosfrac{x}{2}=0\ x=2pi+pi$