Question: solve cosa+sin2a+cos3a=0.
expand the trig terms.
sina = sina
sin2a = 2sina.cosa
sin3a = sin2a.cosa + cos2a.sina = 4*sin(a)*cos(a)^2 - sin(a)
combining these terms.
sina + 2sina.cosa + 4*sin(a)*cos(a)^2 - sin(a) = 0
sina{1 + 2cosa + 4*cos(a)^2 - 1} = 0
sina{2cosa + 4*cos(a)^2} = 0
2sina,cosa{1 + 2*cosa} = 0
sin2a{1 + 2cosa} = 0
Hence, sin2a = 0, 1 + 2cosa = 0
sin2a = 0
2a = nπ, n = 0, ±1, ±2, ...
a = nπ/2, n = 0, ±1, ±2, ...
1 + 2cosa = 0
cosa = -1/2
a = 2π/3 + nπ, n = 0, ±1, ±2, ...
Full solution: a = nπ/2, a = 2π/3 + nπ, n = 0, ±1, ±2, ...