Question

1. Solve for x in the interval [0 , 2 π ).
(a) 2sin^2x=sin x
(b) 3sin^2x+cos^2x=2
(c) 1+3cosx=cos2x
(d) sin2x+sin x=0

Answer 1

a. divide by sinx and will get 2sinx=1 so sinx=1/2
b. here i think that you need using formula sin2(x) + cos2(x) =1
c. check formula for cos2x i think that is 2sinxcosx
d. check formula for sin2x

Answer 2

then i will i break it down,the number 2

Answer 3

have solved number one