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Mathematics 25 Online
OpenStudy (anonymous):

Solve the following: 2cosx+sin(2x)=0 Step by Step explanation please

OpenStudy (anonymous):

\[2\cos(x)+\sin(2x)=0\]you want to work it out together?

OpenStudy (anonymous):

At first, there is a rule,\[\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)\]in your case, however, the "a" and "b" are both "x", so you get, \[\sin(2x)=\sin(x+x)=\sin(x)\cos(x)+\sin(x)\cos(x)=2\sin(x)\cos(x)\]So you can re-write your question as,\[2\cos(x)+2\sin(x)\cos(x)=0\]factor out of 2cos(x), and you get,\[2\cos x(\sin x)=0\] So either,\[2\cos(x)=0~~~~~~~~~or~~~~~~~~~~\sin(x)=0\]

OpenStudy (anonymous):

you should be able to do it from here for the least part.

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