The number of values of in [0,2] for which 2^3 - 7^2+7=2, is:

Question

The number of values of $\alpha$ in $[0,2\pi]$ for which $2\sin^3\alpha - 7\sin^2\alpha+7\sin\alpha=2$, is:

Accepted Answer

$$2\sin^3\alpha-7\sin^2\alpha+7\sin\alpha-2=0$$

Factoring (with $\sin\alpha=1$ as one root): $(\sin\alpha-1)(2\sin^2\alpha-5\sin\alpha+2)=0$

$$\sin\alpha=1 \quad\text{or}\quad \sin\alpha = \frac{5\pm3}4 = \frac12 \text{ or } 2\ (\text{rejected})$$

$\sin\alpha=1 \implies \alpha=\tfrac\pi2$. $\sin\alpha=\tfrac12 \implies \alpha=\tfrac\pi6,\tfrac{5\pi}6$. Total: 3 values.

Solutions · 1