Number of roots of the equation {cos ^2}x + f | vedclass

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Question

$1 - sin^2x + \frac{{\sqrt 3 \, + 1}}{2} \,sinx - \frac{{\sqrt 3 }}{4} - 1 = 0$

$sin^2x -\frac{{\sqrt 3 \, + 1}}{2} \,sinx

Vedclass · Accepted Answer

$4$

Vedclass · Answer

$1 - sin^2x + \frac{{\sqrt 3 \, + 1}}{2} \,sinx - \frac{{\sqrt 3 }}{4} - 1 = 0$

$sin^2x -\frac{{\sqrt 3 \, + 1}}{2} \,sinx + \frac{{\sqrt 3 }}{4} = 0$

$4 sin^2x - 2 \sqrt 3 sinx - 2sinx + \sqrt 3  = 0$

On solving we get

$sinx = 1/2 ; \frac{{\sqrt 3 }}{2}$

         = $(\pi /6 , 5\pi /6 ; \pi /3 , 2\pi /3 ]$

Number of roots of the equation ${\cos ^2}x + \frac{{\sqrt 3 + 1}}{2}\sin x - \frac{{\sqrt 3 }}{4} - 1 = 0$ which lie in the interval $[-\pi,\pi ]$ is

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