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Q.
The number of values of $x$ in the interval $[0, 3\pi]$ satisfying the equation $2sin^2x + 5sinx - 3 = 0$ is
Trigonometric Functions
Solution:
$2sin^{2}\,x+5\,sinx-3=0$
$\Rightarrow sin\,x=\frac{1}{2}$, as $sin\,x \ne-3$
For $0^{\circ} \le x \le 540^{\circ}$, the values of $x$ are $30^{\circ}$, $150^{\circ}, 390^{\circ}, 510^{\circ}$.
$\therefore $ Number of values are $4$.