Question

Find all solutions to tan(x+π)+2sin(x+π)=0 on [0,2π).

Answer 1

To solve the equation \( \tan(x + \pi) + 2\sin(x + \pi) = 0 \) on the interval \( [0, 2\pi) \), we will first utilize the periodic properties of the tangent and sine functions.

We know that:

1. \( \tan(x + \pi) = \tan x \)
2. \( \sin(x + \pi) = -\sin x \)

Substituting these into our equation, we get:

\[
\tan x - 2\sin x = 0
\]

This can be rewritten as:

\[
\tan x = 2\sin x
\]

Using the definition of tangent, we substitute \( \tan x = \frac{\sin x}{\cos x} \) into the equation:

\[
\frac{\sin x}{\cos x} = 2\sin x
\]

Assuming \(\sin x \neq 0\), we can divide both sides by \(\sin x\):

\[
\frac{1}{\cos x} = 2
\]

This simplifies to:

\[
\cos x = \frac{1}{2}
\]

The cosine function equals \( \frac{1}{2} \) at:

\[
x = \frac{\pi}{3}, \quad x = \frac{5\pi}{3}
\]

Next, we need to check the case when \( \sin x = 0 \):

\[
\sin x = 0 \Rightarrow x = 0, \pi
\]

Thus, we have the candidate solutions:

\[
x = 0, \quad x = \frac{\pi}{3}, \quad x = \pi, \quad x = \frac{5\pi}{3}
\]

Finally, we will confirm that all of our candidate solutions satisfy the original equation:

1. For \( x = 0 \):
\[
\tan(0 + \pi) + 2\sin(0 + \pi) = \tan \pi + 2\sin \pi = 0 + 0 = 0
\]

2. For \( x = \frac{\pi}{3} \):
\[
\tan\left(\frac{\pi}{3} + \pi\right) + 2\sin\left(\frac{\pi}{3} + \pi\right) = \tan\left(\frac{4\pi}{3}\right) + 2\sin\left(\frac{4\pi}{3}\right) = \tan\left(\frac{4\pi}{3}\right) + 2\left(-\frac{\sqrt{3}}{2}\right)
\]
\[
= \frac{1}{\sqrt{3}} - \sqrt{3} = \frac{1 - 3}{\sqrt{3}} = \frac{-2}{\sqrt{3}} \neq 0
\]

3. For \( x = \pi \):
\[
\tan(\pi + \pi) + 2\sin(\pi + \pi) = \tan(2\pi) + 2\sin(2\pi) = 0 + 0 = 0
\]

4. For \( x = \frac{5\pi}{3} \):
\[
\tan\left(\frac{5\pi}{3} + \pi\right) + 2\sin\left(\frac{5\pi}{3} + \pi\right) = \tan\left(\frac{8\pi}{3}\right) + 2\sin\left(\frac{8\pi}{3}\right) = \tan\left(\frac{2\pi}{3}\right) + 2\left(-\frac{\sqrt{3}}{2}\right)
\]
\[
= -\frac{1}{\sqrt{3}} - \sqrt{3} = \frac{-1 - 3}{\sqrt{3}} = \frac{-4}{\sqrt{3}} \neq 0
\]

Thus, the valid solutions to the equation \( \tan(x + \pi) + 2\sin(x + \pi) = 0 \) within the interval \( [0, 2\pi) \) are:

\[
\boxed{0, \pi}
\]