The number of solutions of the equation (4-√3) x-2 √3 ^2 x=-41+√3, x [-2 , (5 )/(2)] is

Question

The number of solutions of the equation

$(4-\sqrt{3}) \sin x-2 \sqrt{3} \cos ^2 x=-\frac{4}{1+\sqrt{3}}, x \in\left[-2 \pi, \frac{5 \pi}{2}\right]$ is

Accepted Answer

To find the number of solutions to the equation

$ (4-\sqrt{3}) \sin x - 2 \sqrt{3} \cos^2 x = -\frac{4}{1+\sqrt{3}}, \quad x \in \left[-2\pi, \frac{5\pi}{2}ight] $

we start by letting $\sin x = t$, which implies $\cos^2 x = 1 - t^2$. Substituting these into the equation, we have:

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

Simplifying, we get:

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

Further simplification yields:

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

This simplifies to:

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

To find $t$, solve the quadratic equation:

$ t = \frac{(\sqrt{3}-4) \pm \sqrt{19 - 8 \sqrt{3} + 8(2\sqrt{3})}}{4 \sqrt{3}} $

This simplifies to:

$ t = \frac{(\sqrt{3}-4) \pm \sqrt{19 + 8 \sqrt{3}}}{4 \sqrt{3}} = \frac{(\sqrt{3}-4) \pm (\sqrt{3}+4)}{4 \sqrt{3}} $

Evaluating further, we find:

$ t = \frac{2 \sqrt{3}}{4 \sqrt{3}} \quad 	ext{or} \quad \frac{-8}{4 \sqrt{3}} $

This gives us $ \sin x = \frac{1}{2} $

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