Given that \( \sin x=\cos \left(2 x-27^{\circ}\right) \), find \( \tan x \), correct to 4 significant figures.

To solve the equation \( \sin x = \cos(2x - 27^\circ) \), we can use the identity that relates sine and cosine: \[ \sin x = \cos(90^\circ - x) \] This means we can rewrite the equation as: \[ \cos(90^\circ - x) = \cos(2x - 27^\circ) \] Using the property of cosine, we know that if \( \cos A = \cos B \), then: \[ A = B + 360^\circ n \quad \text{or} \quad A = -B + 360^\circ n \] for some integer \( n \). Applying this to our equation, we have two cases to consider: ### Case 1: \[ 90^\circ - x = 2x - 27^\circ + 360^\circ n \] Rearranging gives: \[ 90^\circ + 27^\circ = 2x + x + 360^\circ n \] \[ 117^\circ = 3x + 360^\circ n \] \[ 3x = 117^\circ - 360^\circ n \] \[ x = \frac{117^\circ - 360^\circ n}{3} \] ### Case 2: \[ 90^\circ - x = - (2x - 27^\circ) + 360^\circ n \] Rearranging gives: \[ 90^\circ - x = -2x + 27^\circ + 360^\circ n \] \[ 90^\circ - 27^\circ = -2x + x + 360^\circ n \] \[ 63^\circ = -x + 360^\circ n \] \[ x = 360^\circ n - 63^\circ \] Now, we will consider the simplest case where \( n = 0 \) for both cases. ### From Case 1: \[ x = \frac{117^\circ}{3} = 39^\circ \] ### From Case 2: \[ x = -63^\circ \quad \text{(not in the range of interest)} \] Now we have \( x = 39^\circ \). Next, we need to find \( \tan x \): \[ \tan 39^\circ \] Using a calculator, we find: \[ \tan 39^\circ \approx 0.8098 \] Thus, the value of \( \tan x \), correct to 4 significant figures, is: \[ \boxed{0.8098} \]