(a) 证明方程

可以写成

(b) 由此，解方程

其中 $0^\circ\leq \theta \leq 360^\circ$。

用 $\cos\theta/\sin\theta$ 代替 $1/\tan\theta$，再乘以 $\sin\theta$。最后利用平方恒等式消去 $\sin^2\theta$。

On the domain of the original equation, both $\sin\theta$ and $\cos\theta$ are nonzero. Thus $1/\tan\theta=\cos\theta/\sin\theta$, and multiplication by $\sin\theta$ gives

Substituting $\sin^2\theta=1-\cos^2\theta$ and rearranging gives

承接 (a)，先把方程看作关于 $\cos\theta$ 的二次方程。两个余弦值分别为正和负，再用一周内的对称性找齐角度。

From (a),

so $\cos\theta=\frac23$ or $\cos\theta=-\frac12$. For $\cos\theta=\frac23$, the principal angle is $\alpha=\cos^{-1}(2/3)$ and the second angle is $360^\circ-\alpha$. For $\cos\theta=-\frac12$, the reference angle is $60^\circ$, giving $120^\circ$ and $240^\circ$.

All four angles have nonzero sine and cosine, so the solutions are