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True/False? The function f(x)=3cos(3x)-2sin(x) has at least one root in the interval [0,pi/2}.
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Just do f(0) and f(pi/2)
if the signs are opposite then you win
first evaluate f(0) then f(pi/2)....so you get f(0)=3cos(0)-2sin(0)=3-0=3 and f(pi/2)=0-2=-2...so your function is continuous on the closed interval, so according to theorem it takes all the values between 0 and pi/2 so at some point it must have value zero, since it changes to negative to positive...
if I didn't mention true :D
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