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solve the following equation for x where 0

Mathematics
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hint: \(sin2x = 2sinxcosx\)
yes and
well where are you stuck or show me what you did

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Other answers:

the equation is not solved
\[ 2sinxcosx = cosx => 2sinxcosx - cosx = 0 \] why dont you try and solve it
take cos(x) you will have \[\Large \cos(x)[2\sin(x)-1]=0\] now you have two equations \[\Large \implies \cos(x)=0\] \[\Large \implies 2\sin(x)-1=0\] solve these two equations.
if i knew how to solve these equations i wouldn;t be asking for help
to this stage i know, but how to solve from here?
\[\cos(x) = 0 => cosx = \cos\left(\frac{\pi}{2}\right) => x = \frac{\pi}{2} +n*2\pi , n \epsilon\mathbb{Z} \] \[sinx = \sin\left(\frac{\pi}{6}\right) => x = \left(-1\right)^{n}*\frac{\pi}{6} +n*\pi , n \epsilon\mathbb{Z} \]
i've been to wolfram alpha to mimi
lol i didnt use wolfram..
the answer is correct though but i have to show it using the unit circle, or the graph
i think using the sin wave
|dw:1345619287591:dw|
ok let me give you an example then if you have 2cos(x)+1=0 then \[\Large \implies \cos(x)=\frac{-1}{2}\] \[\Large x=\cos^{-1}(\frac{-1}{2})\] there are only two points in 9o.2pi) for which cos(x) is -1/2 they are \[\Large x=\frac{2 \pi}{3}\] \[\Large x=\frac{4 \pi}{3}\] now solve the 2sin(x)-1=0 its quite similar to this.
for sinx |dw:1345619404355:dw|
you use wolfram,thats why you are unable to solve these equations by yourself. this wolfram thingy has ruined the students :(. i think at this level students should avoid using that.
yeah i agree. you should only use it to check your answers
no i dont use it but i am aware of it
mimi you've been great i think i should become a fan

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