How do you solve $2sinthetacostheta+sqrt3sintheta=0$ ?

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Answer 1 · 2016-12-17T14:33:06

$theta=npi$ or $theta=2npi+-(5pi)/6$ , where $n$ is an integer

$2sinthetacostheta+sqrt3sintheta=0$

$hArrsintheta(2costheta+sqrt3)=0$

i.e. either $sintheta=0$ and $theta=npi$ , where $n$ is an integer

or $2costheta+sqrt3=0$ and $costheta=-sqrt3/2=cos(+-(5pi)/6)$

and $theta=2npi+-(5pi)/6$ , where $n$ is an integer

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