2sin^(2)x-sinx=1 Find the general form and all solutions in the interval [0,2pi) ?

Feb 27, 2018

$x = k \pi + {\left(- 1\right)}^{k} \frac{\pi}{2} , k \in Z$ or
$x = k \pi - {\left(- 1\right)}^{k} \frac{\pi}{6} , k \in Z$

Explanation:

$2 {\sin}^{2} x - \sin x = 1$
$2 {\sin}^{2} x - 2 \sin x + \sin x - 1 = 0$
2sinx(sinx-1)+1(sinx-1)=0
(sinx-1)(2sinx+1)=0

$\sin x = 1 \mathmr{and} \sin x = - \frac{1}{2}$
$\sin x = \sin \left(\frac{\pi}{2}\right) \mathmr{and} \sin x = \sin \left(- \frac{\pi}{6}\right)$
So,general solution :
$x = k \pi + {\left(- 1\right)}^{k} \frac{\pi}{2} , k \in Z$ or
$x = k \pi - {\left(- 1\right)}^{k} \frac{\pi}{6} , k \in Z$
If ,$x \in \left[0 , 2 \pi\right]$ then
$\sin x = 1 \Rightarrow x = \frac{\pi}{2} ,$
$\sin x = \left(- \frac{1}{2}\right) = \sin \left(- \frac{\pi}{6}\right) \Rightarrow x = \frac{7 \pi}{6} , \frac{11 \pi}{6}$
where, $\frac{\pi}{2} , \frac{7 \pi}{6} , \frac{11 \pi}{6} \in \left[0.2 \pi\right]$
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