Sin3x+cos7x=0 <=> cos(π/2-3x)+cos7x=0 <=> 2*cos(π/2-3x+7x)/2*cos(π/2-3x-7x)/2=0.
Қосындысы 0-ге тең, онда көпмүшеліктің біріде 0-ге тең,яғни:
а) cos(π/4+2x)=0 <=> π/4+2x=π/2+π*n <=> 2x=π/4+π*n <=> x=π/8+π*n/2 n∈Z
б)cos(π/4-5x)=0 <=> π/4-5x=π/2+π*s <=> -5x=π/4+π*s <=> x=-π/20-π*s/5 s∈Z
