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

1 answer

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:

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