\(\theta = 30^\circ, 90^\circ, 150^\circ, 270^\circ\).

February 22, 2026

$\(\theta = 30^\circ, 90^\circ, 150^\circ, 270^\circ\).$

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\theta = 90^\circ \quad \text{and} \quad \theta = 270^\circ.
Checking these values in the original equation, we find \(\sin(180^\circ) = 0\) and \(\sin(540^\circ) = 0\), both equal to \(\cos(90^\circ) = 0\) and \(\cos(270^\circ) = 0\), so they are valid.
Thus, the complete set of solutions is:
\boxed{30^\circ, 90^\circ, 150^\circ, 270^\circ}
Find the vector \(\mathbf{v}\) such that \(\mathbf{v} \times \mathbf{b} = \langle 1, 0, -1 \rangle\) and \(\mathbf{b} = \langle 0, 1, 0 \rangle\).
The vector cross product \(\mathbf{v} \times \mathbf{b} = \langle v_1, v_2, v_3 \rangle \times \langle 0, 1, 0 \rangle\) is computed as:

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