Find the principal solutions of 2sin2x−sin x−1=0

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Question

Find the principal solutions of $2 s i n^{2} x - s i n x - 1 = 0$

Options:

A .

x = \frac{π}{6}

B .

x = \frac{π}{2}

C .

x = - \frac{π}{6}

D .

x = - \frac{π}{2}

Answer: Option B
:
B and C

$2 s i n^{2} x - s i n x - 1 = 0 L e t s i n x = t 2 t^{2} - t - 1 = 0 t = \frac{1 \pm \sqrt{1 - 4 \times 2 \times (- 1)}}{2 \times 2} t = \frac{1 \pm 3}{4} t = 1 o r t = - \frac{1}{2} s i n x = 1 o r s i n x = - \frac{1}{2} x = \frac{π}{2} o r x = - \frac{π}{6}$

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