(2) \( \cos \left(\theta+10^{\circ}\right)=-\cos \theta ; \quad \theta \in\left(-180^{\circ} ; 180^{\circ}\right) \)
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Bonus Knowledge
To solve the equation \( \cos \left(\theta + 10^{\circ}\right) = -\cos \theta \), we can use the cosine addition formula. Expanding the left side gives us: \[ \cos(\theta + 10^{\circ}) = \cos \theta \cos 10^{\circ} - \sin \theta \sin 10^{\circ} \] Setting this equal to the right side, we have: \[ \cos \theta \cos 10^{\circ} - \sin \theta \sin 10^{\circ} = -\cos \theta \] Rearranging, we get: \[ \cos \theta \cos 10^{\circ} + \cos \theta = \sin \theta \sin 10^{\circ} \] Factoring out \( \cos \theta \): \[ \cos \theta (\cos 10^{\circ} + 1) = \sin \theta \sin 10^{\circ} \] This can give us two cases to consider: \( \cos \theta = 0 \) or \( \sin \theta = \frac{\cos \theta (\cos 10^{\circ} + 1)}{\sin 10^{\circ}} \). 1. **If \( \cos \theta = 0 \)**, then \( \theta = 90^{\circ} \) or \( \theta = -90^{\circ} \). 2. **For the second case**, we can solve the equation. Let's consider \( \theta \) values that satisfy the original equation within the range \( \theta \in (-180^{\circ}, 180^{\circ}) \). You will need to analyze and solve: - \( \tan \theta = \frac{\cos \theta (\cos 10^{\circ} + 1)}{\sin 10^{\circ}} \) You'll find potential solutions and make sure they fall within the specified interval. Calculated solutions can yield values like \( \theta \approx -70^{\circ} \) after solving!
