Question

please help! (:

Solve sin2x + 2sinx = 0 for 0 ≤ x < 2 π .

Answer 1

{0}

{0, π/2}

{0, π} <<<<

{π/2,3π/2 }

Answer 2

@kc_kennylau

Answer 3

Simple

Answer 4

Now here we will have 2 equation cosx +1 =0 cosx=-1 x=pi sinx=0 x=pi or 0

Answer 5

\[\sin2x+2sinx=0 \rightarrow 2sinx.cosx+2sinx=0 \rightarrow 2sinx(cosx+1)=0\]
then sinx=0 or cosx+1=0
\[sinx=0 \rightarrow x=k \pi \]
\[cosx+1=0 \rightarrow cosx=-1 \rightarrow x=k \pi + \pi\]
then the answers are \[\left\{ 0, \pi \right\}\]