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How do you solve $2 {sin}^{2} θ - sin θ = 0$ $2sin^2 theta-sin theta=0$ ?

2 Answers

Apr 18, 2016

$θ = n \cdot π$ $theta = n*pi$ or
$θ = \frac{π}{6} + n \cdot 2 π$ $theta= pi/6+n*2pi$ or
$θ = \frac{5 π}{6} + n \cdot 2 π$ $theta=(5pi)/6 +n*2pi$
$color(white)("XXXXXXXXXXXXXXXX")nin ZZ$

Explanation:

If $2sin^2theta-sin theta = 0$
then
$color(white)("XXX")(sin theta)(2sin theta -1)= 0$

$rarr color(white)("XXX")sin theta=0color(white)("XXXXX")orcolor(white)("XXXXX")2sin theta =1$
$color(white)("XXXX")rarr theta = 0+n*pi color(white)("XXXXXXXX")rarr theta=pi/6+2npi or (5pi)/6+2npi$

Apr 18, 2016

I tried to solve it as a normal quadratic equation in $sintheta$ instead of $x$ :

Explanation:

I would collect $sintheta$ and write:

$sintheta[2sintheta-1]=0$

so:

$sintheta=0$ when $theta=0,pi,2pi...$ or $theta=npi$ with $n=0,1,2,3,4....$

and:

$2sintheta-1=0$
$sintheta=1/2$ when $theta=pi/6,5/6pi....$ and multiples of $2npi$ each.

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