Asked by Anonymous
What are the exact solutions within the interval [0,2pi)?
sin2x=cosx
sin2x=cosx
Answers
Answered by
Steve
2sinx cosx = cosx
cosx(2sinx-1) = 0
so, where do you have
cosx = 0
sinx = 1/2
??
cosx(2sinx-1) = 0
so, where do you have
cosx = 0
sinx = 1/2
??
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