Question
Solve the following equations:
$\sin\text{x}-3\sin2\text{x}+\sin3\text{x}=\cos\text{x}-3\cos2\text{x}+\cos3\text{x}$
$\sin\text{x}-3\sin2\text{x}+\sin3\text{x}=\cos\text{x}-3\cos2\text{x}+\cos3\text{x}$
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$\tan(\text{2x}+1)$
$\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{n}-1)\text{d})=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$