6.2 \( [\sin (-\theta)+\cos (360-\theta)]\left[\cos (90-\theta)+\frac{\tan \theta}{\tan }\right] \) If \( \tan x=m+\frac{1}{m}, 90^{\circ} \leq x \leq 270^{\circ} \) and \( m^{2}+\frac{1}{m^{2}}=1 \) Calculate the value of \( x \) without the use of a calculator.

To solve the equation \( \tan x = m + \frac{1}{m} \) under the condition \( m^2 + \frac{1}{m^2} = 1 \), we first manipulate the second equation. Recalling that \( m^2 + \frac{1}{m^2} = (m + \frac{1}{m})^2 - 2 \), we set: \[ (m + \frac{1}{m})^2 - 2 = 1 \] This implies: \[ (m + \frac{1}{m})^2 = 3 \] Taking the square root gives: \[ m + \frac{1}{m} = \pm \sqrt{3} \] Now, solving for \( m \): 1. If \( m + \frac{1}{m} = \sqrt{3} \), then the equation \( m^2 - \sqrt{3}m + 1 = 0 \) can be solved using the quadratic formula: \[ m = \frac{\sqrt{3} \pm i}{2} \] Here, \( m \) is complex, which we will discard for now. 2. If \( m + \frac{1}{m} = -\sqrt{3} \), then again using the quadratic formula: \[ m^2 + \sqrt{3}m + 1 = 0 \] The discriminant is: \[ \Delta = (\sqrt{3})^2 - 4 = 3 - 4 = -1 \] This will also yield complex \( m \). Since neither case yielded \( m \) as a real number, we know then that: Considering \( x \) in the range \( 90^{\circ} \leq x \leq 270^{\circ} \), we know that this covers the second and third quadrants where the tangent values are negative. The only angle that satisfies \( \tan x = -\sqrt{3} \) in this range is: \[ x = 120^{\circ} \text{ (for } \sqrt{3} \text{)} \, \text{and} \, x = 240^{\circ} \text{ (for } -\sqrt{3} \text{)}. \] Thus, considering the properties of \( \tan x \) within defines quadrants, we finally confirm: \[ x = 240^{\circ}. \] So, the value of \( x \) is: \[ \boxed{240^\circ}. \]