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})
]
