\[ \cos 75^\circ = \cos(45^\circ + 30^\circ) \]

\[ \cos(45^\circ + 30^\circ) = \cos 45^\circ\,\cos 30^\circ \;-\; \sin 45^\circ\,\sin 30^\circ \]

\[ = \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} \;-\; \frac{\sqrt2}{2}\cdot\frac{1}{2} = \frac{\sqrt6-\sqrt2}{4}. \]

\[ \cos 15^\circ = \cos(45^\circ - 30^\circ) \]

\[ \cos(45^\circ - 30^\circ) = \cos 45^\circ\,\cos 30^\circ \;+\; \sin 45^\circ\,\sin 30^\circ \]

\[ = \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} \;+\; \frac{\sqrt2}{2}\cdot\frac{1}{2} = \frac{\sqrt6+\sqrt2}{4}. \]

\[ \tan 105^\circ = \tan(60^\circ + 45^\circ) \]

\[ \tan(60^\circ + 45^\circ) = \frac{\tan 60^\circ + \tan 45^\circ}{\,1 - \tan 60^\circ\,\tan 45^\circ\,} = \frac{\sqrt3 + 1}{\,1 - \sqrt3\,} \]

\[ = -\,\frac{\sqrt3 + 1}{\sqrt3 - 1} = -\,\frac{(\sqrt3 + 1)^2}{(\sqrt3 - 1)(\sqrt3 + 1)} = -\,\frac{3 + 2\sqrt3 + 1}{2} = -\,(2+\sqrt3). \]

\[ \sin 15^\circ = \sin(45^\circ - 30^\circ) \]

\[ = \sin 45^\circ\,\cos 30^\circ \;-\; \cos 45^\circ\,\sin 30^\circ = \frac{\sqrt6-\sqrt2}{4}. \]

\[ \frac{1+\sin 15^\circ}{1-\sin 15^\circ} = \frac{\,1+\frac{\sqrt6-\sqrt2}{4}\,}{\,1-\frac{\sqrt6-\sqrt2}{4}\,} = \frac{\,4+\sqrt6-\sqrt2\,}{\,4-\sqrt6+\sqrt2\,}. \]

\[ \sin\frac{5\pi}{12} = \sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right) \]

\[ = \sin\frac{\pi}{4}\cos\frac{\pi}{6} + \cos\frac{\pi}{4}\sin\frac{\pi}{6} \]

\[ = \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} + \frac{\sqrt2}{2}\cdot\frac12 = \frac{\sqrt2}{4}(\sqrt3+1). \]

\[ \cos\frac{7\pi}{12} = \cos\left(\frac{\pi}{3} + \frac{\pi}{4}\right) \]

\[ = \cos\frac{\pi}{3}\cos\frac{\pi}{4} - \sin\frac{\pi}{3}\sin\frac{\pi}{4} \]

\[ = \frac12 \cdot \frac{\sqrt2}{2} - \frac{\sqrt3}{2} \cdot \frac{\sqrt2}{2} = \frac{\sqrt2}{4}(1 - \sqrt3). \]

\[ \tan\frac{\pi}{12} = \tan\left(\frac{\pi}{3} - \frac{\pi}{4}\right) \]

\[ = \frac{\tan\frac{\pi}{3} - \tan\frac{\pi}{4}} {1 + \tan\frac{\pi}{3}\tan\frac{\pi}{4}} = \frac{\sqrt3 - 1}{1 + \sqrt3 \times 1} = \frac{\sqrt3 - 1}{\sqrt3 + 1}. \]

Multiply top and bottom by \(\sqrt3 - 1\):

\[ = \frac{(\sqrt3 - 1)^2}{(\sqrt3 + 1)(\sqrt3 - 1)} = \frac{3 - 2\sqrt3 + 1}{3 - 1} = \frac{4 - 2\sqrt3}{2} = 2 - \sqrt3. \]

\[ \sin\left(-\frac{\pi}{12}\right) = \sin\left(\frac{\pi}{4} - \frac{\pi}{3}\right) \]

\[ = \sin\frac{\pi}{4}\cos\frac{\pi}{3} - \cos\frac{\pi}{4}\sin\frac{\pi}{3} \]

\[ = \frac{\sqrt2}{2}\cdot\frac12 - \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} = \frac{\sqrt2}{4}(1-\sqrt3). \]

Use the identity \(\displaystyle \sin(A+B)=\sin A\cos B+\cos A\sin B\).

First find the cosines from right-triangle ratios:
\[ \cos A=\sqrt{1-\sin^2A} =\sqrt{1-\left(\frac{4}{5}\right)^2} =\sqrt{\frac{9}{25}}=\frac{3}{5}, \] \[ \cos B=\sqrt{1-\sin^2B} =\sqrt{1-\left(\frac{5}{13}\right)^2} =\sqrt{\frac{144}{169}}=\frac{12}{13}. \]

Substitute into the compound-angle formula:
\[ \sin(A+B) =\left(\frac{4}{5}\right)\!\left(\frac{12}{13}\right) +\left(\frac{3}{5}\right)\!\left(\frac{5}{13}\right) =\frac{48}{65}+\frac{15}{65} =\frac{63}{65}. \]

Use the compound-angle identity \[ \tan(45^\circ-\theta)= \frac{\tan45^\circ-\tan\theta}{1+\tan45^\circ\,\tan\theta} =\frac{1-\tan\theta}{1+\tan\theta}. \]

Substitute into the equation and simplify: \[ \tan\theta =2\left(\frac{1-\tan\theta}{1+\tan\theta}\right) \quad\Rightarrow\quad \tan\theta(1+\tan\theta)=2(1-\tan\theta). \] \[ \tan^2\theta+3\tan\theta-2=0. \]

Let \(t=\tan\theta\). Then \[ t=\frac{-3\pm\sqrt{(-3)^2-4(1)(-2)}}{2} =\frac{-3\pm\sqrt{17}}{2} \approx 0.5615\;\text{ or }\;-3.5615. \]

Convert to angles in \(0^\circ\le\theta\le360^\circ\):
\(\tan\theta\approx0.5615\Rightarrow \theta\approx29.3^\circ,\,209.3^\circ.\)
\(\tan\theta\approx-3.5615\Rightarrow \theta\approx105.7^\circ,\,285.7^\circ\) (opposite quadrants for negative tangent).

Expand both sides using compound-angle formulae:

\[ \cos(\theta+45^\circ) =\cos\theta\cos45^\circ-\sin\theta\sin45^\circ =\frac{\sqrt2}{2}\big(\cos\theta-\sin\theta\big), \] \[ \sin(\theta-45^\circ) =\sin\theta\cos45^\circ-\cos\theta\sin45^\circ =\frac{\sqrt2}{2}\big(\sin\theta-\cos\theta\big) =-\frac{\sqrt2}{2}\big(\cos\theta-\sin\theta\big). \]

Substitute into the equation and simplify (split across lines for readability):

\[ 5\cdot\frac{\sqrt2}{2}\,(\cos\theta-\sin\theta) =-\frac{\sqrt2}{2}\,(\cos\theta-\sin\theta) \;\Longrightarrow\; 5(\cos\theta-\sin\theta)=-(\cos\theta-\sin\theta). \]

\[ 6(\cos\theta-\sin\theta)=0 \;\Longrightarrow\; \cos\theta-\sin\theta=0 \;\Longrightarrow\; \tan\theta=1. \]

Within \(0^\circ\le\theta\le360^\circ\): \(\theta=45^\circ,\,225^\circ\).

Expand: \[ \cos(x+30^\circ)=\cos x\cos30^\circ-\sin x\sin30^\circ =\tfrac{\sqrt3}{2}\cos x-\tfrac12\sin x. \] Set equal to \(2\sin x\): \[ \tfrac{\sqrt3}{2}\cos x=\tfrac52\sin x \;\Rightarrow\; \tan x=\frac{\sqrt3}{5}. \]

Answer: \(x=19.1^\circ,\,199.1^\circ\).

Expand: \[ \cos(x-60^\circ)=\cos x\cos60^\circ+\sin x\sin60^\circ =\tfrac12\cos x+\tfrac{\sqrt3}{2}\sin x. \] Set equal to \(3\cos x\): \[ \tfrac{\sqrt3}{2}\sin x=\tfrac52\cos x \;\Rightarrow\; \tan x=\frac{5}{\sqrt3}. \] \(x=70.9^\circ,\,250.9^\circ\).

Expand: \[ \sin(30^\circ-x)=\tfrac12\cos x-\tfrac{\sqrt3}{2}\sin x. \] Set equal to \(4\sin x\): \[ \tfrac12\cos x=\Big(4+\tfrac{\sqrt3}{2}\Big)\sin x \Rightarrow \tan x=\frac{1}{8+\sqrt3}. \]

Answer: \(x=5.9^\circ,\,185.9^\circ\).

Expand both sides: \[ \tfrac{\sqrt3}{2}\cos x-\tfrac12\sin x =\sin x+\sqrt3\cos x. \] Rearranging gives \[ \sqrt3\cos x+3\sin x=0 \Rightarrow \tan x=-\frac{\sqrt3}{3}. \]

Answer: \(x=150^\circ,\,330^\circ\).

Let \(t=\tan x\). Using \(\displaystyle \tan(60^\circ-x)=\frac{\sqrt3-t}{1+\sqrt3\,t}\): \[ 2\frac{\sqrt3-t}{1+\sqrt3 t}=t \Rightarrow \sqrt3 t^2+3t-2\sqrt3=0. \] Divide by \(\sqrt3\): \(t^2+\sqrt3\,t-2=0\).

Answer: \(x=38.4^\circ,\,111.6^\circ\).

With \(t=\tan x\) and \(\displaystyle \tan(45^\circ-x)=\frac{1-t}{1+t}\): \[ 2\frac{1-t}{1+t}=3t\Rightarrow 3t^2+5t-2=0. \]

Answer: \(x=18.4^\circ,\,116.6^\circ\).

Expand and collect: \[ \tfrac12\sin x+\tfrac{\sqrt3}{2}\cos x =\sqrt2\cos x-\sqrt2\sin x. \] Hence \[ \big(\tfrac12+\sqrt2\big)\sin x +\big(\tfrac{\sqrt3}{2}-\sqrt2\big)\cos x=0 \Rightarrow \tan x=\frac{0.286\ldots}{}. \]

Answer: \(x=16.0^\circ\).

Expand: \[ \tfrac{\sqrt2}{2}(\sin x-\cos x)=\cos x-\sqrt3\sin x. \] Rearranging gives \[ (\sqrt2+2\sqrt3)\sin x-(\sqrt2+2)\cos x=0 \Rightarrow \tan x=\frac{\sqrt2+2}{\sqrt2+2\sqrt3}. \]

Answer: \(x=35.0^\circ\).

With \(t=\tan x\) and \(\displaystyle \tan(x+45^\circ)=\frac{t+1}{1-t}\): \[ \frac{t+1}{1-t}=6t \Rightarrow 6t^2-5t+1=0 \Rightarrow t=\frac12,\;\frac13. \]

Answer: \(x=18.4^\circ,\,26.6^\circ\).

Expand: \[ -\tfrac{\sqrt2}{2}\cos x+\tfrac{\sqrt2}{2}\sin x =\sin x-\sqrt3\cos x. \] Hence \[ \big(\tfrac{\sqrt2}{2}-1\big)\sin x +\big(-\tfrac{\sqrt2}{2}+\sqrt3\big)\cos x=0 \Rightarrow \tan x\approx 3.498. \]

Answer: \(x=74.1^\circ\).

Let \(t=\tan x\). Using \(\displaystyle \tan(x-45^\circ)=\frac{t-1}{1+t}\) and \(\cot x=\frac1t\): \[ \frac{t-1}{1+t}+\frac1t=2 \Rightarrow \frac{t^2+1}{t(1+t)}=2 \Rightarrow t^2+2t-1=0. \] Thus \(t=-1\pm\sqrt2\).

Answer: \(x=22.5^\circ,\,112.5^\circ\).