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:

The given equation is $$2\sin^3\alpha - 7\sin^2\alpha + 7\sin\alpha = 2$$, and we need to find the number of values of $$\alpha$$ in the interval $$[0, 2\pi]$$ that satisfy it. First, rearrange the equation to bring all terms to one side: $$$ 2\sin^3\alpha - 7\sin^2\alpha + 7\sin\alpha - 2 = 0 $$$ To simplify, substitute $$x = \sin\alpha$$. Since $$\alpha$$ is in $$[0, 2\pi]$$, $$x$$ ranges from $$-1$$ to $$1$$. The equation becomes: $$$ 2x^3 - 7x^2 + 7x - 2 = 0 $$$ This is a cubic equation. Using the rational root theorem, possible rational roots are $$\pm 1, \pm 2, \pm \frac{1}{2}$$. Testing $$x = 1$$: $$$ 2(1)^3 - 7(1)^2 + 7(1) - 2 = 2 - 7 + 7 - 2 = 0 $$$ So, $$x = 1$$ is a root. Factor out $$(x - 1)$$ using synthetic division. The coefficients are $$2, -7, 7, -2$$. Dividing by $$(x - 1)$$: - Bring down 2. - Multiply by 1: $$2 \times 1 = 2$$. - Add to next coefficient: $$-7 + 2 = -5$$. - Multiply by 1: $$-5 \times 1 = -5$$. - Add to next coefficient: $$7 + (-5) = 2$$. - Multiply by 1: $$2 \times 1 = 2$$. - Add to last coefficient: $$-2 + 2 = 0$$. The quotient is $$2x^2 - 5x + 2$$. So, the cubic factors as: $$$ (x - 1)(2x^2 - 5x + 2) = 0 $$$ Set each factor to zero: - $$x - 1 = 0$$ gives $$x = 1$$. - $$2x^2 - 5x + 2 = 0$$. Solve using the quadratic formula: $$$ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{4} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4} $$$ So, $$x = \frac{8}{4} = 2$$ or $$x = \frac{2}{4} = \frac{1}{2}$$. The roots are $$x = 1$$, $$x = 2$$, and $$x = \frac{1}{2}$$. Since $$x = \sin\alpha$$ must be in $$[-1, 1]$$, $$x = 2$$ is invalid. The valid solutions are $$x = 1$$ and $$x = \frac{1}{2}$$. Now, solve for $$\alpha$$ in $$[0, 2\pi]$$: - $$\sin\alpha = 1$$ has one solution: $$\alpha = \frac{\pi}{2}$$. - $$\sin\alpha = \frac{1}{2}$$ has two solutions: $$\alpha = \frac{\pi}{6}$$ and $$\alpha = \frac{5\pi}{6}$$. Verify each solution in the original equation: - For $$\alpha = \frac{\pi}{6}$$, $$\sin\alpha = \frac{1}{2}$$: $$$ 2\left(\frac{1}{2}\right)^3 - 7\left(\frac{1}{2}\right)^2 + 7\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{8} - 7 \cdot \frac{1}{4} + 7 \cdot \frac{1}{2} = \frac{1}{4} - \frac{7}{4} + \frac{14}{4} = \frac{1 - 7 + 14}{4} = \frac{8}{4} = 2 $$$ - For $$\alpha = \frac{\pi}{2}$$, $$\sin\alpha = 1$$: $$$ 2(1)^3 - 7(1)^2 + 7(1) = 2 - 7 + 7 = 2 $$$ - For $$\alpha = \frac{5\pi}{6}$$, $$\sin\alpha = \frac{1}{2}$$: $$$ 2\left(\frac{1}{2}\right)^3 - 7\left(\frac{1}{2}\right)^2 + 7\left(\frac{1}{2}\right) = \frac{1}{4} - \frac{7}{4} + \frac{14}{4} = \frac{8}{4} = 2 $$$ All three solutions satisfy the equation. The solutions are distinct and within $$[0, 2\pi]$$: $$\alpha = \frac{\pi}{6}$$, $$\alpha = \frac{\pi}{2}$$, and $$\alpha = \frac{5\pi}{6}$$. Hence, the number of values of $$\alpha$$ is 3. So, the answer is 3.