What is tanα equal to?

Calculation: 

We are given:

\( 2\sin(\alpha) + \cos(\alpha) = 2 \)

\( \cos(\alpha) = 2(1 - \sin(\alpha)) \)

\( \cos^2(\alpha) = 4(1 - \sin(\alpha))^2 \)

Use the identity \(\cos^2(\alpha) = 1 - \sin^2(\alpha) \)

\( 1 - \sin^2(\alpha) = 4(1 - 2\sin(\alpha) + \sin^2(\alpha)) \)

\( 1 - \sin^2(\alpha) = 4 - 8\sin(\alpha) + 4\sin^2(\alpha) \)

Rearrange the terms to form a quadratic equation:

\( 5\sin^2(\alpha) - 8\sin(\alpha) + 3 = 0 \)

Using the quadratic formula:

\( \sin(\alpha) = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(5)(3)}}{2(5)} \)

\( \sin(\alpha) = \frac{8 \pm \sqrt{64 - 60}}{10} \)

\( \sin(\alpha) = \frac{8 \pm 2}{10} \)

\( \sin(\alpha) = 1 \) or \( \sin(\alpha) = \frac{3}{5} \)

Since \(0^\circ < \alpha < 90^\circ \), we select \(\sin(\alpha) = \frac{3}{5} \)

\( \cos^2(\alpha) = 1 - \sin^2(\alpha) = 1 - \left( \frac{3}{5} \right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \)

\( \cos(\alpha) = \frac{4}{5} \)

\( \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \)

∴ The correct answer is Option (c):