20.3. sin 2z + cos 2z = √2*sin 3z
Есть формула
sin a + cos a = √2*(1/√2*sin a + 1/√2*cos a) =
= √2*(sin a*cos pi/4 + cos a*sin pi/4) = √2*sin(a + pi/4)
Подставляем
√2*sin (pi/4 + 2z) = √2*sin 3z
sin(pi/4 + 2z) = sin 3z
sin(pi/4 + 2z) - sin 3z = 0
2sin [(pi/4 + 2z - 3z)/2]*cos [(pi/4 + 2z + 3z)/2] = 0
2sin (pi/8 - z/2)*cos (pi/8 + 5z/2) = 0
1) sin (pi/8 - z/2) = 0; pi/8 - z/2 = pi*k; z1 = pi/4 + 2pi*k
2) cos (pi/8 + 5z/2) = 0;
pi/8 + 5z/2 = pi/2 + 2pi*n; z2 = 3pi/20 + 4pi/5*n\
pi/8 + 5z/2 = 3pi/2 + 2pi*m; z3 = 11pi/20 + 4pi/5*m