The number of values of x in the interval [0,3] satisfying the equation 2^2x+5 x-3=0 is:

Question

The number of values of $x$ in the interval $[0,3\pi]$ satisfying the equation $2\sin^2x+5\sin x-3=0$ is:

Accepted Answer

$$(\sin x+3)(2\sin x-1)=0 \implies \sin x=\frac12\ (\sin x=-3 \text{ rejected})$$

In $[0,3\pi]$: $x = \dfrac\pi6,\dfrac{5\pi}6,\ 2\pi+\dfrac\pi6,\ 2\pi+\dfrac{5\pi}6$ --- 4 values.

Solutions · 1