i) \( \left[\operatorname{sen}(x+2 \pi)+\cos \left(\frac{\pi}{2}-x\right)\right]^{2}+\frac{4 \cos (x-2 \pi)}{\csc \left(\frac{\pi}{2}-x\right)}=4 \)

To solve the equation given: \[ \left[\operatorname{sen}(x+2 \pi)+\cos \left(\frac{\pi}{2}-x\right)\right]^{2}+\frac{4 \cos (x-2 \pi)}{\csc \left(\frac{\pi}{2}-x\right)}=4 \] we start by simplifying the terms involved: 1. The function \( \operatorname{sen}(x) \) represents the sine function, which can be expressed as \( \sin(x) \). Also, since sine and cosine functions have a periodicity of \( 2\pi \), we have: \[ \operatorname{sen}(x + 2\pi) = \sin(x) \] and \[ \cos(x - 2\pi) = \cos(x). \] 2. The term \( \cos\left(\frac{\pi}{2}-x\right) \) simplifies to \( \sin(x) \) because of the co-function identity: \[ \cos\left(\frac{\pi}{2}-x\right) = \sin(x). \] 3. The term \( \csc\left(\frac{\pi}{2}-x\right) \) is the reciprocal of the sine function: \[ \csc\left(\frac{\pi}{2}-x\right) = \sec(x). \] Rewriting the expression using these simplifications: \[ [\sin(x) + \sin(x)]^{2} + \frac{4 \cos(x)}{\sec(x)} = 4. \] This equation can be further simplified: \[ (2\sin(x))^{2} + 4\cos(x) \cdot \cos(x) = 4. \] Thus: \[ 4\sin^2(x) + 4\cos^2(x) = 4. \] We can factor out the \( 4 \): \[ 4(\sin^2(x) + \cos^2(x)) = 4. \] Using the Pythagorean identity \( \sin^2(x) + \cos^2(x) = 1 \): \[ 4(1) = 4, \] which holds true. Thus, the original equation is satisfied by any \( x \), which indicates that the solution is: **All real numbers \( x \)**. Therefore, the solution to the equation is all values of \( x \).