Question

find the general solution of sin 2x+cos x = 1

Answer 1

sin2x = 1-cosx
2sinxcosx = 2(sin(x/2))^2
4sin(x/2) cos(x/2)cosx - 2(sin(x/2))^2 =0
2sin(x/2) ( 2cos(x/2)cosx - sin(x/2) ) =0
we may get one solution from 2sin(x/2) =0
or x/2 = n pi
or x=2n pi and further...

Answer 2

one solution is given as (2n+1)pi/2 and another is n(pi) + (-1)^n 7pi / 6
how do i get these ? @matricked

Answer 3

is your question with sin2x or (sin x)^2

Answer 4

is is sin 2 x as in sin (two x )

Answer 5

i think it means (sin x)^2
well if it is so
then
we can rewrite the question as
(sin x)^2 +cosx =1
or
1-(cos x)^2x +cosx =1

Answer 6

it is not sin ^2 x but sin (2x) + cos x = 1

Answer 7

by the way if we put n=0 in the answer provided by you
the solutions are two of the solutions pi/2 and pi/6
but none of them satisfies sin (2x) + cos x = 1

Answer 8

i meant 7 pi/6

Answer 9

can we proceed like this 2 sin x cos x + cos x = 1
or cos x (2 sin x+1)= 1
hence either cos x =1 or 2sin x +1 = 1 ?

Answer 10

no you can't

Answer 11

as a* 1/a = 1 for all a<>0