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solve the following equation for x where
0<or = to x<2pi, sin2x=cosx
 one year ago
 one year ago
solve the following equation for x where 0<or = to x<2pi, sin2x=cosx
 one year ago
 one year ago

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Mimi_x3Best ResponseYou've already chosen the best response.3
hint: \(sin2x = 2sinxcosx\)
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.3
well where are you stuck or show me what you did
 one year ago

jpjonesBest ResponseYou've already chosen the best response.0
the equation is not solved
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.3
\[ 2sinxcosx = cosx => 2sinxcosx  cosx = 0 \] why dont you try and solve it
 one year ago

sami21Best ResponseYou've already chosen the best response.1
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.
 one year ago

jpjonesBest ResponseYou've already chosen the best response.0
if i knew how to solve these equations i wouldn;t be asking for help
 one year ago

jpjonesBest ResponseYou've already chosen the best response.0
to this stage i know, but how to solve from here?
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.3
\[\cos(x) = 0 => cosx = \cos\left(\frac{\pi}{2}\right) => x = \frac{\pi}{2} +n*2\pi , n \epsilon\mathbb{Z} \] \[sinx = \sin\left(\frac{\pi}{6}\right) => x = \left(1\right)^{n}*\frac{\pi}{6} +n*\pi , n \epsilon\mathbb{Z} \]
 one year ago

jpjonesBest ResponseYou've already chosen the best response.0
i've been to wolfram alpha to mimi
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.3
lol i didnt use wolfram..
 one year ago

jpjonesBest ResponseYou've already chosen the best response.0
the answer is correct though but i have to show it using the unit circle, or the graph
 one year ago

jpjonesBest ResponseYou've already chosen the best response.0
i think using the sin wave
 one year ago

sami21Best ResponseYou've already chosen the best response.1
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.
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.3
for sinx dw:1345619404355:dw
 one year ago

sami21Best ResponseYou've already chosen the best response.1
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.
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.3
yeah i agree. you should only use it to check your answers
 one year ago

jpjonesBest ResponseYou've already chosen the best response.0
no i dont use it but i am aware of it
 one year ago

jpjonesBest ResponseYou've already chosen the best response.0
mimi you've been great i think i should become a fan
 one year ago
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