How do you solve #6sin(x/2)cos(x/2)+3cos(x/2)-4sin(x/2)-2=0#?

1 Answer
Mar 10, 2017

#pi/3 + 4kpi; (5pi)/3 + 4kpi#
#96^@38 + k720^@; 263^@62 + k720^@#

Explanation:

Factor by grouping:
#3cos (x/2)(2sin (x /2) + 1) - 2(2sin (x/2) + 1) = 0#
#(2sin (x/2) + 1)(3cos (x/2) - 2) = 0#
Use trig table and unit circle -->
a. #2sin (x/2) + 1 = 0# --> #sin (x/2) = - 1/2# --> 2 solutions -->
#x/2 = (7pi)/6) + 2kpi# --> #x = (7pi)/3 + 4kpi = pi/3 + 4kpi#
#x = (11pi)/6 + 2kpi # --> #x = (11pi)/3 + 4kpi = (5pi/3) = 4kpi#
b. #3cos (x/2) - 2 = 0 #--> #cos (x/2) = 2/3#
Use calculator and unit circle -->
#cos (x/2) = 2/3# --> #x/2 = +- 48^@19# --> 2 solutions -->
#x/2 = 48^@19 + k360^@# --> #x = 96^@38 + k720^@#
#x/2 = 360 - 48.19 = 311^@81 + k360^@# -->
#x = 623^@62 + k720^@ = 263^@62 + k720^@#