sinx + 2cosx = 0

To solve the equation \( \sin x + 2\cos x = 0 \), we can rearrange it as follows:

\[ \sin x = -2\cos x \]

Next, we can divide both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)) to get:

\[ \tan x = -2 \]

Now, to find the general solutions for \( x \), we take the arctangent:

\[ x = \tan^{-1}(-2) + n\pi \]

where \( n \) is any integer, since the tangent function has a period of \( \pi \).

To find a specific solution, we can compute \( \tan^{-1}(-2) \), which is approximately:

\[ x \approx -1.107 \text{ radians} \]

Thus, the general solution for the equation \( \sin x + 2\cos x = 0 \) is:

\[ x \approx -1.107 + n\pi, \quad n \in \mathbb{Z} \]

You can also express the solution in degrees if needed, taking into consideration that \( -1.107 \text{ radians} \approx -63.43^\circ \). The angle in degrees would be:

\[ x \approx -63.43^\circ + n \cdot 180^\circ \]

for \( n \in \mathbb{Z} \).