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8sin^3(x) will always be less or equal than 8 and greater than or equal to -8, since sine function will always generate less than 7cos2x will always be less than or equal to 7 and greater than or equal to -7, so, if p < -15 then it will not have any solution and also if p > 15, same is the case.
But I've given the answer and it's supposed to be p=(-inf;-15)u(7;inf)
How do you get the 7 to infinity part?
Got it. Thanks!
(-inf;-15)u(15;inf) is my answer
Yes, mine must be wrong.
incase cos part is -7 and the sin part is 8, than p must be 15, if p is greater than 15, then no solution exist.
Oo... wait, maximize and minimize the following function, you will get a lot better answer f(x) = 8sin^3(x)-7cos2x
How do you do that?
take first derivative ... and equate it to zero. and solve for x. for now!! and alert me
Okay. I'll let you know.
x=nPi , x=2nPi , x=(-1)^n+nPi
Second one is wrong.
let's see http://www.wolframalpha.com/input/?i=+y+%3D+8sin%5E3%28x%29-7cos2x+
Yes, got those already.
in these points you either get maximum value or minimum value
the best way would be to put these values in our function and check which gives maximum value and which gives minimum value
beyond greatest maximum value, solution does not exist .... and same goes to minimum value
Alright. I try to write it all down. Thank you!
and your answer is p<-7 and p>15
thats from graph