Given \( 5 \tan x+4=0 \) and \( x \in\left[180^{\circ} ; 360^{\circ}\right] \). Use a suitable diagram to determine the following, without using a calculator: \( \sin ^{2}\left(x-90^{\circ}\right)+\sin ^{2} x \)

To solve the equation \( 5 \tan x + 4 = 0 \), we simplify it to find \( \tan x = -\frac{4}{5} \). Since \( x \) is restricted to the interval \([180^{\circ}, 360^{\circ}]\), we know that tangent is negative in the third and fourth quadrants. In this case, \( x \) must lie in the fourth quadrant. Thus, to determine \( x \), we can consider the reference angle where \( \tan^{-1}\left(\frac{4}{5}\right) \), which can be called \( \theta \), satisfies the property that \( \tan \theta = \frac{4}{5} \). In the fourth quadrant, we calculate \( x \) as: \[ x = 360^{\circ} - \theta \] Now to find \( \sin^2(x - 90^\circ) + \sin^2 x \): 1. \( \sin(x - 90^\circ) = \cos x \). 2. Hence, \( \sin^2(x - 90^\circ) = \cos^2 x \). We can express the desired quantity as: \[ \sin^2(x - 90^\circ) + \sin^2 x = \cos^2 x + \sin^2 x = 1 \] Therefore, the answer is: \[ \sin^2(x - 90^\circ) + \sin^2 x = 1. \]