Given that 6 cos (x+15)^circ-6 sin (2 x-30)^circ=0 find tan (2 x-10^circ) (caving wan answer in ard form

To solve the given equation \(6 \cos(x+15)^{\circ} - 6 \sin(2x-30)^{\circ} = 0\), we can simplify it by dividing both sides by 6:\(\cos(x+15)^{\circ} - \sin(2x-30)^{\circ} = 0\)Now, we can use the trigonometric identity \(\sin(2x-30)^{\circ} = \sin(2x)\cos(30^{\circ}) - \cos(2x)\sin(30^{\circ})\):\(\cos(x+15)^{\circ} - (\sin(2x)\cos(30^{\circ}) - \cos(2x)\sin(30^{\circ})) = 0\)Simplifying further, we get:\(\cos(x+15)^{\circ} - \sin(2x)\cos(30^{\circ}) + \cos(2x)\sin(30^{\circ}) = 0\)Now, we can use the trigonometric identity \(\cos(x+15)^{\circ} = \cos(x)\cos(15^{\circ}) - \sin(x)\sin(15^{\circ})\):\(\cos(x)\cos(15^{\circ}) - \sin(x)\sin(15^{\circ}) - \sin(2x)\cos(30^{\circ}) + \cos(2x)\sin(30^{\circ}) = 0\)Now, we can use the trigonometric identity \(\cos(15^{\circ}) = \frac{\sqrt{6} + \sqrt{2}}{4}\) and \(\sin(15^{\circ}) = \frac{\sqrt{6} - \sqrt{2}}{4}\):\(\cos(x)\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) - \sin(x)\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) - \sin(2x)\cos(30^{\circ}) + \cos(2x)\sin(30^{\circ}) = 0\)Now, we can use the trigonometric identity \(\cos(30^{\circ}) = \frac{\sqrt{3}}{2}\) and \(\sin(30^{\circ}) = \frac{1}{2}\):\(\cos(x)\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) - \sin(x)\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) - \sin(2x)\left(\frac{\sqrt{3}}{2}\right) + \cos(2x)\left(\frac{1}{2}\right) = 0\)Now, we can use the trigonometric identity \(\sin(2x) = 2\sin(x)\cos(x)\) and \(\cos(2x) = \cos^2(x) - \sin^2(x)\):\(\cos(x)\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) - \sin(x)\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) - 2\sin(x)\cos(x)\left(\frac{\sqrt{3}}{2}\right) + \left(\cos^2(x) - \sin^2(x)\right)\left(\frac{1}{2}\right) = 0\)Now, we can use the trigonometric identity \(\cos^2(x) + \sin^2(x) = 1\):\(\cos(x)\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) - \sin(x)\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) - 2\sin(x)\cos(x)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right) = 0\)Now, we can use the trigonometric identity \(\cos(x) = \frac{1 + \cos(2x)}{2}\) and \(\sin(x) = \frac{\sqrt{1 - \cos^2(x)}}{2}\):\(\frac{1 + \cos(2x)}{2}\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) - \frac{\sqrt{1 - \cos^2(x)}}{2}\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) - 2\left(\frac{\sqrt{1 - \cos^2(x)}}