Updated on August 6, 2022
Find all solutions of the equation cos x (2sin x + 1) = 0.
The answer is A+kpi, B+2kpi and C+2kpi where k is any integer and 0<A<pi, 0<B<C<2pi,
the answer asks for
A=?
B=?
C=?
I figured out that A=pi/2
but what does B and C equal? if B is B+2kpi and same for C
so the answer for B was 7pi/6 so Im still trying to figure out C
4 Answers

cosx(2sinx + 1) = 0 ⇒ (1) cosx = 0 or (2) 2sinx + 1 = 0
(1) cosx = 0 ⇒ x = π/2 + kπ
(2) 2sinx + 1 = 0 ⇒ sinx = 1/2
. . . . . . . . . . . .. ⇒ x = π/6 + 2kπ or x = π – (π/6) + 2kπ = 7π/6 + 2kπ

cos(x)(2sin(x) + 1) = 0
so cos(x) = 0 or sin(x) = 1/2
cos(x) = 0 when x = +/ Ï/2 + 2kÏ
sin(x) = 1/2 when x = Ï/6 + 2kÏ or 7Ï/6 + 2kÏ
will leave it to you to convert Ï/2 and Ï/6 to the proper range

0<A<pi
x = pi/2
0<B<C<2pi,
x = pi/2  x = (7 pi)/6  x = (3 pi)/2  x = (11 pi)/6

This link has a step by step solution to your problem:
http://www.symbolab.com/math/solver/stepbystep/t…
Hope this helps