First slide

Trigonometric equations

Question

Number of integral values of ‘a’ satisfy the equation ${[\sin x]}^{2} + \sin x - 2 a = 0$ is (where [.] denotes G.I.F)

Moderate

A

1
B

2
C

3
D

4

Solution

$W e h a v e 2 a = {[\sin x]}^{2} + \sin x$
$\begin{array}{l} ∵ a \in I \Rightarrow \sin x \in I \\ \Rightarrow [\sin x] = \sin x \end{array}$
$∴ 2 a = \sin x (\sin x + 1)$ and $\sin x = - 1 o r 0 o r 1$
$\Rightarrow 2 a = 0 o r 2$ $\Rightarrow a = 0 o r 1$

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