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Mathematics 19 Online
OpenStudy (anonymous):

csc^2x+2cscx=0

OpenStudy (anonymous):

what's csc?

OpenStudy (campbell_st):

its the abbreviation for cosec so looking at the problem you have a quadratic equation... factoring and you get csc(x)(csc(x) + 2) = 0 so you need to find the values of x where csc(x) = 0 and csc(x) = 2 I think one of the solutions doesn't exist.. hope it helps

OpenStudy (anonymous):

Solve the given equation over the interval [0, 2π): cos2 x = cos x. a)3pi/2 b)pi/2 c)2pi d)0 sorry didn't write out problem

OpenStudy (anonymous):

^different problem lol

OpenStudy (campbell_st):

is it cos(2x) or cos^2(2x)

OpenStudy (anonymous):

cos^2(x)=cos(x)

OpenStudy (campbell_st):

but it looks like another quadratic rewriting it and you get \[\cos^2(x) - \cos(x) =0\] factoring and you get cos(x)(cos(x) - 1) = 0 so just solve for x again

OpenStudy (anonymous):

k thanks

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