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If (cos 2x)(sin 2x)=0, what's the best approach for solving for x? I can see that it will be zero at any point where x=0, but is there a way I should be simplifying the expression first? I'm (painfully) discovering that I've got a few gaps still from having taught myself trig.
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(cos 2x)(sin 2x)=0 means that either cos 2x=0 or sin 2x=0 or both, so solve both those equations individually. The set of answers that you get between them is the answer to the initial problem.
\[\sin 2x \cos 2x =\frac{ 2 }{ 2 }\sin 2x \cos 2x =\frac{ 1 }{ 2 } \sin 4x =0 \\ \rightarrow \sin 4x=0 \\4x=0 +k \pi\]
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