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sinx(sinx+2cosx) = 0
What can you do now? think.

if a*b = 0,
Either a = 0 or b = 0
Here, a = sinx and b = sinx + 2cosx

i don't think this is that easy

\[sin^2x+2\ sin\ x\ cos\ x=0\]
\[sin\ x(sin\ x+2\ cos\ x)=0\]
\[(sin\ x+2\ cos\ x)=0\]

that is not to say you cannot do it, but it is going to take some work

in general you can write
\[a\sin(x)+b\cos(x)\] as \[\sqrt{a^2+b^2}\sin(x+\theta)\]

:'( i need help on mine

I kinda wonder if you could do \(\sin^2(x) + \sin(2x)=0\).

in this case you get
\[\sin(x)+2\cos(x)=\sqrt{5}\sin(x+\theta)\] where \(\tan(\theta)=2\)

tanx = -2
x = atan(-2) + n*(pi)

therefore \(x+\theta=n\pi\) gives \(\theta=n\pi-\tan^{-1}(2)\)

oh I had forgotten this site Thanx @satellite73 for remembering :)

Gaahh @satellite73 you were right.