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Given the function f (x) = the square of 2acosx + bsinxcosx root sign 3 / 2, and f (0) = root sign 3 / 2, f (PAI / 4) = 1 / 2, find the analytic expression of F (x), simple increasing interval
Because f (0) = √ 3 / 2
So a = √ 3 / 2
Because f (π / 4) = 1 / 2
So B = 1
So f (x) = 2 √ 3 / 2cos ^ x + sinxcosx - √ 3 / 2
=2√3/2[(cos2x+1)/2]+1/2sin2x-√3/2
=sin(2x+60)
Simple increasing interval (K π - 75, K π + 15)
Simple subtraction interval (K π + 15, K π + 105)
The function f (x) = ACOS square x + bsinxcosx satisfies f (0) = 2, f (3 / 3) = 1 / 2 + 2 / 2 root 3, and finds the value of A.B
a=b=2,
The known function f (x) = 2acos ^ 2x + bsinxcosx - 1
And f (0) = 1, f (π / 3) = - 1 / 2 + [(radical 3) / 2]
1. Find the value of a and B
2. Find monotone decreasing interval
3. Find the set of F (x) > 0, X
F (0) = 2a-1 = 1 a = 1 f (π / 3) = 2 * (1 / 4) + b * [(radical 3) / 2] * (1 / 2) - 1 = - 1 / 2 + [(radical 3) / 2] B = 22. F (x) = cos2x + sin2x = (radical 2) sin [2x + (π / 4)] period is π (π / 8) + K π ≤ x ≤ (5 π / 8) + K π 3. Sin [2x + (π / 4)] > 00 < 2x + (π / 4) < π (- π / 8) + K π
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