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if \[\log_{\sin x}\cos x-1+\log_{\sin x}\sin 2x \] then the values of x are
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i assume that + means = then cot x = sin2x sinx = +1/root2 , -1/root2
oh crap! \[\log_{\sin x}\cos x=1+\log_{\sin x}\sin 2x\] this is the Q
although it doesnt differ much from what you assumed=)
did u get it now??
how did you simplify the log terms bro??
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write 1 as log(sinx) to base sinx
then bring it to the other side..it becomes log(cotx) to base sinx
ok got it..no need to xplain nemore buddy=)
thnx
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