4 (1+cosx) = 3 sin^2x/2cosx/2

4 (1+cosx) = 3sin²x/2 сosx/2

cos²x/2=1+cosx/2 |*2

2cos²x/2=1+cosx

4 (2cos²x/2) = 3sin²x/2*cosx/2

8cos²x/2-3sin²x/2*cosx/2=0

cosx/2 (8cosx/2-3sin²x/2) = 0

cosx/2=0

x/2=п/2+пn, n€z

x=п+2 пn, n€z

8cosx/2-3sin²x/2=0

8cosx/2-3 (1-cos²x/2) = 0

3cos²x/2+8cosx/2-3=0

cosx/2=t

3t²+8t-3=0

D=64-4*3 * (-3) = 100

t1=-8+10/6=1/3

t2=-8-10/6=-18/6=-3

cosx/2=1/3

x/2=+-arccos1/3+2 пn, n€z

x=+-2arccos1/3+4 пn, n€z

cosx/2=-3

нет решения.

Ответ: x=п+2 пn, n€z; x=+-2arccos1/3+4 пn, n€z