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anonymous
 3 years ago
solve the following equation for x where
0<or = to x<2pi, sin2x=cosx
anonymous
 3 years ago
solve the following equation for x where 0<or = to x<2pi, sin2x=cosx

This Question is Closed

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.3hint: \(sin2x = 2sinxcosx\)

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.3well where are you stuck or show me what you did

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the equation is not solved

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.3\[ 2sinxcosx = cosx => 2sinxcosx  cosx = 0 \] why dont you try and solve it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0take 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if i knew how to solve these equations i wouldn;t be asking for help

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0to this stage i know, but how to solve from here?

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.3\[\cos(x) = 0 => cosx = \cos\left(\frac{\pi}{2}\right) => x = \frac{\pi}{2} +n*2\pi , n \epsilon\mathbb{Z} \] \[sinx = \sin\left(\frac{\pi}{6}\right) => x = \left(1\right)^{n}*\frac{\pi}{6} +n*\pi , n \epsilon\mathbb{Z} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i've been to wolfram alpha to mimi

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.3lol i didnt use wolfram..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the answer is correct though but i have to show it using the unit circle, or the graph

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think using the sin wave

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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.

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.3for sinx dw:1345619404355:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you 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.

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.3yeah i agree. you should only use it to check your answers

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no i dont use it but i am aware of it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0mimi you've been great i think i should become a fan
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