(i) Express the equation \(\sin 2x + 3 \cos 2x = 3(\sin 2x - \cos 2x)\) in the form \(\tan 2x = k\), where \(k\) is a constant.
(ii) Hence solve the equation for \(-90^\circ \leq x \leq 90^\circ\).
Solution
(i) Start with the equation:
\(\sin 2x + 3 \cos 2x = 3(\sin 2x - \cos 2x)\)
Expand the right side:
\(\sin 2x + 3 \cos 2x = 3\sin 2x - 3\cos 2x\)
Rearrange terms:
\(\sin 2x + 3 \cos 2x - 3\sin 2x + 3\cos 2x = 0\)
\(-2\sin 2x + 6\cos 2x = 0\)
Divide by \(\cos 2x\):
\(-2\tan 2x + 6 = 0\)
\(\tan 2x = 3\)
Thus, \(k = 3\).
(ii) Solve \(\tan 2x = 3\) for \(-90^\circ \leq x \leq 90^\circ\):
\(2x = \tan^{-1}(3)\)
\(2x = 71.6^\circ\) or \(2x = -108.4^\circ\)
Divide by 2:
\(x = 35.8^\circ\) or \(x = -54.2^\circ\)
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