Question

Find all solutions in the interval [0, 2π).

2 sin^2x = sin x

I watched a live lesson, but I'm still completely stuck... lol

Answer 1

\[\large \bf 2\sin^2x=sinx\]
\[\large \bf 2\sin^2x-sinx=0\]
Take `sinx ` common,
\[\large \bf sinx(2sinx-1)=0\]
so either sinx=0 OR (2sinx-1)=0

Answer 2

solve these 2 equations and find solutions in the interval [0,2pi)

Answer 3

hope you understand @zeeke

Answer 4

@mayankdevnani I understand up until the very last part.... How would I solve those? Substitute in a trigonometric identity...? I'm slightly confused.

These are the answer choices btw, I mistakenly left them out.
x = π/3, 2π/3
x = π/2, 3π/2, π/3, 2 π / 3.
x = 0, π, π/six, 5π/6
x = π/6, 5π/6

Answer 5

what you think ?

Answer 6

i think the most appropriate answer would be option C but to correct option C we have to change the interval
\[\large \bf OLD~INTERVAL \rightarrow [0,2 \pi)\]
\[\large \bf NEW~INTERVAL \rightarrow [0,2 \pi]\]
then answer would be option C
otherwise option D is answer

Answer 7

understood ? @zeeke

Answer 8

@mayankdevnani I'm so sorry, but I'm honestly still kinda lost. Where exactly do these intervals come into play? The lesson I took seemed to teach everything BESIDES these interval sort of problems, so I'm completely taken aback by these questions...

Answer 9

I'm just not quite sure of what to do after I know

sinx=0 and (2sinx-1)=0

Answer 10

I'm just not quite sure of what to do after I know

sinx=0 and (2sinx-1)=0