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To solve the equation

\[2 \sin ( 2x) + \sin ( x) = 0\]

for all solutions in the interval \(0 \leq x < 2\pi \), we start by using the double-angle identity for \(\sin ( 2x) \):

\[\sin ( 2x) = 2 \sin ( x) \cos ( x) \]

Substitute this into the original equation:

\[2 \cdot 2 \sin ( x) \cos ( x) + \sin ( x) = 0\]

\[4 \sin ( x) \cos ( x) + \sin ( x) = 0\]

Factor out \(\sin ( x) \):

\[\sin ( x) ( 4 \cos ( x) + 1) = 0\]

This equation is satisfied if either \(\sin ( x) = 0\) or \(4 \cos ( x) + 1 = 0\).

Case 1: \(\sin ( x) = 0\)

\(\sin ( x) = 0\) at:

\[x = 0, \pi , 2\pi \]

Since we are considering the interval \(0 \leq x < 2\pi \), the solutions from this case are:

\[x = 0, \pi \]

Case 2: \(4 \cos ( x) + 1 = 0\)

Solve for \(\cos ( x) \):

\[4 \cos ( x) + 1 = 0\]

\[4 \cos ( x) = - 1\]

\[\cos ( x) = - \frac { 1} { 4} \]

Now, find the values of \(x\) for which \(\cos ( x) = - \frac { 1} { 4} \) within the interval \(0 \leq x < 2\pi \). These values correspond to:

\[x = \cos ^ { - 1} \left ( - \frac { 1} { 4} \right ) \]

\[x = 2\pi - \cos ^ { - 1} \left ( - \frac { 1} { 4} \right ) \]

Using a calculator to find the approximate values:

\[x \approx \cos ^ { - 1} \left ( - \frac { 1} { 4} \right ) \approx 1.8235\]

\[x \approx 2\pi - \cos ^ { - 1} \left ( - \frac { 1} { 4} \right ) \approx 2\pi - 1.8235 \approx 4.4597\]

Final Solutions

Combining the solutions from both cases, we get:

\[x = 0, \pi , 1.82, 4.46\]

Thus, the solutions to \(2 \sin ( 2x) + \sin ( x) = 0\) in the interval \(0 \leq x < 2\pi \) are:

\[x = 0, \pi , 1.82, 4.46\]

These values are accurate to at least two decimal places.

Supplemental Knowledge

To solve trigonometric equations, we often use identities and algebraic manipulation to simplify the equation. In this case, we will use the double-angle identity for sine:

\[\sin ( 2x) = 2\sin ( x) \cos ( x) \]

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