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Finding the maximum and minimum of the function y = - 2cos ^ 2 + 2sinx + 3 / 2
y=2*（1-sin^2)+2sin+3/2=-2sin^2+2sin+3/2
So when sin = 1 / 2, the maximum is 4
When sin = - 1, the minimum value is 1 / 2
The minimum and maximum of function f (x) = 2cos & sup2; x-2sinx-1
f(x)=2cos²x-2sinx-1
=2（1-sin²x）-2sinx-1
=1-2sin²x-2sinx
=3/2-2(sinx+1/2)²
-1=
Find the range of y = (SiNx ^ 2 + 3cosx-4) / (cosx-2)
Like the title,
This is not the answer to this question. Let's have a look
Y = f (x) = (4-3sinx) (4-3cosx) = 16-12sinx-12cosx + 9sinxcosx = 16-12 (SiNx + cosx) + 9sinxcosx, let SiNx + cosx = t, then sinxcosx = (T ^ 2-1) / 2 ∵ x is an acute angle, t = SiNx + cosx = √ 2Sin (x + π / 4), ｜ t ∈ (1, √ 2] y = f (x) = 4-3sinx (4-3cosx) = 16-12t + 9 (t
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- 14. Given f (x) = 2sinx / 2cosx / 2 + 2 (cosx / 2) * 2-1, find the set of all real numbers x that make f (x) + F (x) '= 0 I have worked out the original formula F (x) = 2sinx + cosx f(x)'=2cosx-sinx Finally, SiNx cosx = 0 What is the set of X? And what did I do right? The result is that x is π / 4 + 2 π n, and N is a positive integer,
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