Question

solve for theta if 4 sin^2 theta-2(√3+1)sin theta+√3=0​

Answer 1

Answer:

$\theta = \dfrac{\pi }{3}$ and

$\theta = \dfrac{\pi }{6}$

Step-by-step explanation:

Given that-

$4Sin^{2}\theta -2(\sqrt{3}+1)Sin\theta +\sqrt{3} = 0$

$4Sin^{2}\theta -2\sqrt{3}Sin\theta - 2Sin\theta + \sqrt{3} = 0$

$2Sin\theta(2Sin\theta -\sqrt{3}) -1(2Sin\theta - \sqrt{3} = 0$

$(2Sin\theta - \sqrt{3}) (2Sin\theta - 1) = 0$

If, $2Sin\theta - \sqrt{3} = 0$

$2Sin\theta = \sqrt{3}$

$Sin\theta = \dfrac{\sqrt{3} }{2}$

$Sin\theta = Sin\dfrac{\pi }{3}$

$\theta = \dfrac{\pi }{3}$

If, $2sin\theta - 1 =0$

$2Sin\theta = 1$

$Sin\theta = \dfrac{1}{2}$

$Sin\theta = Sin\dfrac{\pi }{6}$

$\theta = \dfrac{\pi }{6}$