The-solution-set-of-the-equation-4-sin-cos-2-cos-2-3-sin-3-0-in-the-interval-0-2pi-is-

Question Number 113986 by deepraj123 last updated on 16/Sep/20

The solution set of the equation

4 \sin θ \cos θ - 2 \cos θ - 2 \sqrt{3} \sin θ + \sqrt{3} = 0

in the interval (0, 2 π) is

Answered by MJS_new last updated on 16/Sep/20

let t = \tan \frac{θ}{2} \Leftrightarrow θ = 2 \arctan t

- \frac{8 t (t^{2} - 1)}{{(t^{2} + 1)}^{2}} + \frac{2 (t^{2} - 1)}{t^{2} + 1} - \frac{4 \sqrt{3} t}{t^{2} + 1} + \sqrt{3} = 0

(2 + \sqrt{3}) t^{4} - 4 (2 + \sqrt{3}) t^{3} + 2 \sqrt{3} t^{2} + 4 (2 - \sqrt{3}) t - 2 + \sqrt{3} = 0

t^{4} - 4 t^{3} - 2 (3 - 2 \sqrt{3}) t^{2} + 4 (7 - 4 \sqrt{3}) t - 7 + 4 \sqrt{3} = 0

{(t - 2 + \sqrt{3})}^{2} (t + 2 - \sqrt{3}) (t - 2 - \sqrt{3}) = 0

t_{1, 2} = 2 - \sqrt{3}

t_{3} = - 2 + \sqrt{3}

t_{4} = 2 + \sqrt{3}

t = \tan \frac{θ}{2}

\Rightarrow x_{1, 2} = \frac{π}{6} x_{3} = \frac{11 π}{6} x_{4} = \frac{5 π}{6}

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