Given that \( \sin 2 x=\cos \left(x-33^{\circ}\right) \), find \( \tan x \), correct to 4 significant figures.

To solve the equation \( \sin 2x = \cos(x - 33^\circ) \), we can use the identity \( \sin 2x = 2\sin x \cos x \). So the equation becomes: \[ 2\sin x \cos x = \cos(x - 33^\circ) \] Now, we can use the cosine of the difference formula, which states that \( \cos(A - B) = \cos A \cos B + \sin A \sin B \). Applying this to \( \cos(x - 33^\circ) \), we get: \[ 2\sin x \cos x = \cos x \cos 33^\circ + \sin x \sin 33^\circ \] Now, we can divide both sides of the equation by \( \cos x \) (assuming \( \cos x \neq 0 \)) to simplify: \[ 2\sin x = \cos 33^\circ + \tan x \sin 33^\circ \] Now, we can solve for \( \tan x \): \[ 2\sin x - \tan x \sin 33^\circ = \cos 33^\circ \] \[ \tan x \sin 33^\circ = 2\sin x - \cos 33^\circ \] \[ \tan x = \frac{2\sin x - \cos 33^\circ}{\sin 33^\circ} \] To find \( \tan x \), we need to know the values of \( \sin x \) and \( \sin 33^\circ \). However, we don't have a specific value for \( x \), so we can't find a numerical value for \( \tan x \) without additional information. If we assume that \( x \) is such that \( \sin x \) and \( \cos 33^\circ \) are known, we could then calculate \( \tan x \). But without that information, we can't solve for \( \tan x \) numerically. If you have a specific value for \( x \) or if you're looking for a general solution, please provide that information.