(a) Use \(\cos(A+B)=\cos A\cos B-\sin A\sin B\):

\[ \cos 3\theta=\cos(2\theta+\theta) =(\cos2\theta)\cos\theta-(\sin2\theta)\sin\theta . \] With \(\cos2\theta=2\cos^2\theta-1\) and \(\sin2\theta=2\sin\theta\cos\theta\), \[ \cos 3\theta =(2\cos^2\theta-1)\cos\theta-2\sin\theta\cos\theta\sin\theta =4\cos^3\theta-3\cos\theta. \]

(b) Substitute the identity from (a) and \(\cos2\theta=2\cos^2\theta-1\):

\[ (4\cos^3\theta-3\cos\theta)+\cos\theta(2\cos^2\theta-1) =\cos^2\theta . \] \[ \Rightarrow\; 6\cos^3\theta-4\cos\theta=\cos^2\theta \;\Rightarrow\; 6\cos^3\theta-\cos^2\theta-4\cos\theta=0 \] \[ \Rightarrow\; \cos\theta\,(6\cos^2\theta-\cos\theta-4)=0 . \]

Either \(\cos\theta=0\Rightarrow \theta=90^\circ\), or \(6\cos^2\theta-\cos\theta-4=0\). Solving the quadratic in \(c=\cos\theta\): \[ c=\frac{1\pm\sqrt{1+96}}{12} =\frac{1\pm\sqrt{97}}{12} \approx 0.9041\;\text{ or }\;-0.7374. \] Hence \[ \theta=\cos^{-1}(0.9041)\approx 25.3^\circ,\qquad \theta=\cos^{-1}(-0.7374)\approx 137.5^\circ, \] together with \(\theta=90^\circ\) from \(\cos\theta=0\).