Solve this equation using sum to product formulas: cos3x + sin2x - sin6x + cos5x = 0Will mark Brainliest.

Answer:x=(nπ/2)±15x=(2nπ+90)/3x=90-2nπSteps:cos3x+sin2x-sin6x+cos5x=0(cos3x+cos5x)+(sin2x-sin6x)=0equation(1)Use formula:[tex] \cos(c) + \cos(d) = 2 \cos( \frac{c + d}{2} ) . \cos( \frac{c - d}{2} ) [/tex][tex] \sin(c) - \sin(d) = 2 \cos( \frac{c + d}{2} ) . \sin( \frac{c - d}{2} ) [/tex]So in equation (1),if c=3x ,d=5x,C=2x,D=6x》2cos{(c+d)/2}.cos{(c-d)/2}+2cos{(C+D)/2}.sin{(C-D)/2}=0》2cos{(3x+5x)/2}.cos{(3x-5x)/2}+2cos{(2x+6x)/2}.sin{(2x-6x)/2}=0》2cos(4x).cos(-x)+2cos(4x).sin(-2x)=0》2cos(4x)[cos(-x)+sin(-2x)]=0》2cos(4x)[cos(x)-sin(2x)]=01) Either:2cos(4x)=0cos(4x)=0/2cos(4x)=0cos(4x)=cos(90)General solution for such case is:X=2n±aSo,4x=2nπ±90x=(2nπ±90)/4x=(nπ/2)±152) Or:cos(x)-sin(2x)=0cos(x)=sin(2x)General solution for such case is:X=2n±aSo,x=2nπ±(90-2x)x=2nπ±90±2xx±2x=2nπ±90Take +ve sign,x+2x=2nπ+903x=2nπ+90x=(2nπ+90)/3Take -ve sign,x-2x=2nπ-90-x=2nπ-90x=(2nπ-90)/(-1)x=90-2nπ