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isn't that 1?
f'(x) = 2Cosx - 2Sin2x = 0
solve for x: 2Cosx - 2Sin(2x) = 0
x = (1/2) pi (4*n-1), where n is an integer
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the only solution in the given interval is: x = 3pi/2
So plug each endpoint and the solution into the equation: x=0 y= 2Sin0 + Cos(2*0) = 0 + 1 = 1
x = 3pi/2 y = 2Sin(3pi/2) + Cos(3pi) = -3
And finally, 2pi: x = 2pi y = 2Sin(2pi) + Cos(4pi) = 1
Thus the maximum of y is equal to 1, which is none of the above, which is option 1.
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thank you so much! what would be the absolute minimun?
absolute minimum is y=-3, which occurs at x = 3pi/2 (assuming the interval's the same)
oh ok and to find all the solutions for sec4x=2 , i solved it and i got pi/12 but i dont know how to find the other solutions
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