\( 2 \tan 2 x=\tan 128^{\circ} \) and \( x \in\left[-360^{\circ} ; 360^{\circ}\right] \) \( 3 \sin \left(x-40^{\circ}\right)=\cos 50^{\circ} \) and \( x \in\left[-360^{\circ} ; 360^{\circ}\right] \)

To solve the equation \(2 \tan 2x = \tan 128^\circ\), you start by finding the reference angles that correspond to the given tangent value. The tangent function is periodic, which means that solutions may be found by adding or subtracting \(180^\circ\) from the angle. Next, for \(3 \sin(x - 40^\circ) = \cos 50^\circ\), you can find \(\cos 50^\circ\) and then isolate \(\sin(x - 40^\circ)\). Remember that \(\sin\) is also periodic; thus, you'd consider additional angles to accommodate all potential solutions within the specified range. Now, let's get a bit more adventurous with the concepts! Did you know that tangent's periodicity means it repeats every \(180^\circ\)? This smooth little trick allows you to find multiple solutions in a blink! So, for \(2 \tan 2x = \tan 128^\circ\), not only do you find those angles but also explore all the angles of the form \(128^\circ + k \cdot 180^\circ\) where \(k\) is an integer. For the sine equation, consider converting \(\cos 50^\circ\) into sine form using the co-function identity \(\cos 50^\circ = \sin 40^\circ\) (because \(90^\circ - 50^\circ = 40^\circ\)). This neat transformation opens new pathways for solutions that feel downright clever!