Find exactly all solutions to the given equation.
\[
\cos 2 \theta+\cos \theta=0, \quad 0^{\circ} \leq \theta< 360^{\circ}
\]

Step 1 ：Given the equation \(\cos 2 \theta + \cos \theta = 0\), where \(0^\circ \leq \theta < 360^\circ\).

Step 2 ：We can use the double angle identity for cosine to simplify the equation. The double angle identity for cosine is \(\cos 2 \theta = 1 - 2 \sin^2 \theta\).

Step 3 ：Substituting this into our equation gives \(1 - 2 \sin^2 \theta + \cos \theta = 0\).

Step 4 ：We can use the Pythagorean identity, \(\sin^2 \theta + \cos^2 \theta = 1\), to express \(\sin^2 \theta\) in terms of \(\cos \theta\): \(\sin^2 \theta = 1 - \cos^2 \theta\).

Step 5 ：Substituting this into our equation gives \(1 - 2(1 - \cos^2 \theta) + \cos \theta = 0\).

Step 6 ：This simplifies to a quadratic equation in terms of \(\cos \theta\): \(2 \cos^2 \theta - \cos \theta - 1 = 0\).

Step 7 ：Solving this quadratic equation gives the values of \(\cos \theta\) as -1/2 and 1.

Step 8 ：We can then use the inverse cosine function to find the values of \(\theta\), which are \(2\pi/3\) and 0.

Step 9 ：Converting these to degrees gives \(\theta = 120^\circ\) and \(\theta = 0^\circ\).

Step 10 ：However, we need to consider the range of the cosine function. The cosine function has a period of 360^\circ, so we need to add multiples of 360^\circ to our solutions to find all solutions in the range \(0^\circ \leq \theta < 360^\circ\).

Step 11 ：Final Answer: The solutions to the equation \(\cos 2 \theta + \cos \theta = 0, \quad 0^\circ \leq \theta < 360^\circ\) are \(\theta = \boxed{0^\circ}\) and \(\theta = \boxed{120^\circ}\).