Question

2root3 sin^[email protected] - ="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="1b7874685b">[email protected]= 0

Answer 1

Answer:

Step-by-step explanation:

$2\sqrt{3} sin^{2} \alpha -cos\alpha =0\\2\sqrt{3} (1-cos ^2 \alpha )-cos \alpha =0\\2\sqrt{3} -2\sqrt{3} cos^2 \alpha -cos \alpha =0\\2\sqrt{3} cos^2 \alpha +cos \alpha -2\sqrt{3} =0\\cos \alpha =\frac{-1 \pm\sqrt{1^2-4*2\sqrt{3}*(-2\sqrt{3}) } }{2*2\sqrt{3} } \\=\frac{-1 \pm\sqrt{1+48} }{4\sqrt{3} } \\=\frac{-1\pm7}{4\sqrt{3} } \\either~cos \alpha =\frac{6}{4\sqrt{3} }=\frac{\sqrt{3} }{2} \\=cos \frac{\pi }{6} ,cos(2\pi -\frac{\pi }{6} )\\=cos \frac{\pi}{6} ,cos \frac{11\pi }{6}$

$\alpha =2 n\pi+ \frac{\pi }{6} ,2n\pi +\frac{11\pi }{6} (general~solution)$

$or~cos\alpha =-\frac{7}{4\sqrt{3} } \\ \alpha =cos^{-1}( \frac{-7}{4\sqrt{3} } )$