First slide

Trigonometric equations

Question

$T h e n u m b e r o f v a l u e s o f x i n t h e i n t e r v a l [0, 3 π] s a t i s f y i n g t h e e q u a t i o n 2 \sin^{2} x + 5 \sin x - 3 = 0 i s$

Moderate

A

1
B

4
C

6
D

2

Solution

$G i v e n e q u a t i o n i s$ $2 \sin^{2} x + 5 \sin x - 3 = 0$
$\Rightarrow (\sin x + 3) (2 \sin x - 1) = 0$
$\Rightarrow \sin x = \frac{1}{2} (∵ \sin x + 3 \neq 0)$
$S i n c e x \in [o, 3 π], \Rightarrow x = \frac{π}{6}, \frac{5 π}{6}, \frac{13 π}{6}, \frac{17 π}{6}$

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