Start from the identity \(\cos(x-y)=\cos x\cos y+\sin x\sin y\).

Compute \(p^2+q^2\):

\[ p^2+q^2 =(\sin x+\sin y)^2 + (\cos x+\cos y)^2 \] \[ =(\sin^2x+\sin^2y + 2\sin x\sin y) +(\cos^2x+\cos^2y + 2\cos x\cos y) \] \[ =( \sin^2x+\cos^2x ) + ( \sin^2y+\cos^2y ) + 2(\sin x\sin y+\cos x\cos y) \] \[ =1+1+2\cos(x-y) =2+2\cos(x-y). \]

Solve for \(\cos(x-y)\):

\[ \cos(x-y)=\frac{p^2+q^2-2}{2}. \]

Result: \(\displaystyle \boxed{\cos(x-y)=\tfrac{p^2+q^2-2}{2}}\).

(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 \(\cos\theta=0\).

Expand each term using \(\sin(u\pm v)=\sin u\cos v\pm\cos u\sin v\) and \(\cos(u\pm v)=\cos u\cos v\mp\sin u\sin v\):

\[ \sin\!\left(x+\tfrac{\pi}{6}\right) -\sin\!\left(x-\tfrac{\pi}{6}\right) =(\sin x\cos\tfrac{\pi}{6}+\cos x\sin\tfrac{\pi}{6}) -(\sin x\cos\tfrac{\pi}{6}-\cos x\sin\tfrac{\pi}{6}) \] \[ =2\cos x\sin\tfrac{\pi}{6} =2\cos x\cdot\tfrac12 =\cos x. \]

\[ \cos\!\left(x+\tfrac{\pi}{3}\right) -\cos\!\left(x-\tfrac{\pi}{3}\right) =(\cos x\cos\tfrac{\pi}{3}-\sin x\sin\tfrac{\pi}{3}) -(\cos x\cos\tfrac{\pi}{3}+\sin x\sin\tfrac{\pi}{3}) \] \[ =-2\sin x\sin\tfrac{\pi}{3} =-2\sin x\cdot\tfrac{\sqrt3}{2} =-\sqrt3\,\sin x. \]

Equate LHS and RHS: \[ \cos x=-\sqrt3\,\sin x \quad\Rightarrow\quad \tan x=\frac{\sin x}{\cos x}=-\frac{1}{\sqrt3}. \]

(b) Since \(\tan x = -\tfrac{1}{\sqrt3}\) (reference angle \(=\tfrac{\pi}{6}\)) and tangent is negative in quadrants II and IV, \[ x = \pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}, \qquad x = 2\pi-\tfrac{\pi}{6}=\tfrac{11\pi}{6}. \]

Final answers

(a) \(\displaystyle \tan x=-\frac{1}{\sqrt3}\).
(b) \(\displaystyle x=\frac{5\pi}{6},\;\frac{11\pi}{6}\) on \(0\le x\le 2\pi\).

Use the tangent addition formula: \( \tan(x+45^\circ)=\frac{\tan x+1}{\,1-\tan x\,}. \) Also \(\cot x=\dfrac{1}{\tan x}\).

Let \(t=\tan x\). Then \[ \frac{t+1}{1-t}=2\cdot\frac{1}{t} \;\Rightarrow\; t(t+1)=2(1-t) \] \[ \Rightarrow\; t^2+t=2-2t \;\Rightarrow\; t^2+3t-2=0. \]

Solve the quadratic: \[ t=\frac{-3\pm\sqrt{9+8}}{2} =\frac{-3\pm\sqrt{17}}{2} \approx 0.5616\;\text{ or }\;-3.5616. \]

\[ \tan x\approx0.5616 \Rightarrow x\approx \arctan(0.5616)=29.3^\circ, \] \[ \tan x\approx-3.5616 \Rightarrow x\text{ in QII } =180^\circ-\arctan(3.5616)=105.7^\circ. \]

Final answer

\(\boxed{x\approx29.3^\circ,\;105.7^\circ}\)

Expand the left side with \(\cos(u-v)=\cos u\cos v+\sin u\sin v\):

\[ \cos(\theta-60^\circ) =\cos\theta\cos60^\circ+\sin\theta\sin60^\circ =\tfrac12\cos\theta+\tfrac{\sqrt3}{2}\sin\theta. \]

Set equal to \(3\sin\theta\) and rearrange:

\[ \tfrac12\cos\theta+\tfrac{\sqrt3}{2}\sin\theta=3\sin\theta \;\Rightarrow\; \tfrac12\cos\theta=\Big(3-\tfrac{\sqrt3}{2}\Big)\sin\theta. \] \[ \cos\theta=(6-\sqrt3)\,\sin\theta \;\Rightarrow\; \tan\theta=\frac{\sin\theta}{\cos\theta} =\frac{1}{\,6-\sqrt3\,}. \]

Evaluate the principal angle:

\[ \theta=\arctan\!\Big(\frac{1}{6-\sqrt3}\Big) \approx \arctan(0.2343)=13.2^\circ. \]

Tangent is positive in quadrants I and III, so add \(180^\circ\):

\[ \theta=13.2^\circ,\quad 13.2^\circ+180^\circ=193.2^\circ. \]