Trigonometrical Equations
normal

સમીકરણ ${\cos ^2}x + \frac{{\sqrt 3  + 1}}{2}\sin x - \frac{{\sqrt 3 }}{4} - 1 = 0$ ના  $[-\pi,\pi ]$ માં ઉકેલોની સંખ્યા ............. છે 

A

$2$

B

$4$

C

$6$

D

$8$

Solution

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

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

$4 sin^2x – 2 \sqrt 3 sinx – 2sinx + \sqrt 3  = 0$

On solving we get

$sinx = 1/2 ; \frac{{\sqrt 3 }}{2}$

         = $(\pi /6 , 5\pi /6 ; \pi /3 , 2\pi /3 ]$

Standard 11
Mathematics

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