Question

cot 2 theta + 3 cos theta =0 find theta​

Answer 1

${ \cos }^{2} θ\: + 3 \: cosθ+ 3 = 0$

$\frac{ {cos}^{2}θ}{ {sin}^{2}θ} + \frac{3}{sinθ} + 3 = 0$

$\frac{cos^{2}θ\: + 3 \: sin θ + 3 {sin}^{2}θ}{sin^{2}θ} = 0$

${cos}^{2} θ + 3 {sin}^{2} θ + 3sinθ$

$1 + 2 {sin}^{2} θ + 3sinθ = 0$

Put sinθ = x in the above equation we get

${2x}^{2} + 3x + 1 = 0$

Solution of above polynomial is given by:-

$x = \frac{ - 3 + \sqrt{ {3}^{2} - 4 \times 2 \times 1} }{ {2}^{2} }$

$x = \frac{ - 3 + \sqrt{9 - 8} }{4}$

$x = \frac{ - 3 + \sqrt{1} }{4}$

$x = \frac{ - 3 + 1}{4}$

Hence,

$x = - 1 \: or \: x = - \frac{1}{2}$

Hence,

$sinθ = - 1 \: or \: sinθ = - \frac{1}{2}$

$> θ = \frac{(4n \ + 3)\pi}{2} \: or \: θ = \frac{(12n + 7)\pi}{6} =$

$\frac{(12n + 11)\pi}{6}$

Where n = 0,1,2,3,...

$itzshinde$