Math, asked by archanatiwarkhede, 10 months ago

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

Answers

Answered by jitekumar4201
61

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}

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