2cos x+sin x = 0

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

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

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

\[ \tan x = -2 \]

Now we can find the general solutions for \( x \). The tangent function is periodic with a period of \( \pi \), so we can write:

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

where \( n \) is any integer.

To express \( \tan^{-1}(-2) \) in terms of a specific angle, we can find it using a calculator or a table of values. Using the inverse tangent function:

\[ \tan^{-1}(-2) \approx -1.107 \text{ radians} \]

Thus, the solutions can be expressed as:

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

Another way to express the solution in degrees (if needed):

\[ \tan^{-1}(-2) \approx -63.43^\circ \]

Then the solutions in degrees would be:

\[ x \approx -63.43^\circ + n \cdot 180^\circ \quad (n \in \mathbb{Z}) \]

So, the final solution set for the equation \( 2\cos x + \sin x = 0 \) is:

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

or in degrees:

\[ x \approx -63.43^\circ + n \cdot 180^\circ \quad (n \in \mathbb{Z}) \]