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Trigonometrical Equations
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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
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
