The minimum positive period of function y = 3cos & # X-1

The minimum positive period of function y = 3cos & # X-1

y = 3(cosx)^2 - 1
= 3/2 *[2(cosx)^2] - 1
= 3/2 *[cos(2x) + 1] - 1
= 3/2 *cos(2x) + 1/2
Because the minimum positive period of COS (2x) is π, the minimum positive period of this function is also π

Y = 3cos & # 178; x-4cosx + 1, X belongs to [π / 3,2 π / 3] to find the maximum value of function, it must be adopted

Let t = cosx, then - 1 / 2=

The known function f (x) = SiNx cosx - √ 3cos & # 178; X + √ 3 / 2 (x ∈ R)
(1) Finding the minimum positive period of F (x)
(2) Finding monotone interval of F (x)
(3) Find the symmetry axis and center of F (x) image

F (x) = sinxcosx - √ 3cos & # 178; X + √ 3 / 2 (x ∈ R) = 1 / 2 sin2x - √ 3 / 2 (2cos & # 178; x-1) = 1 / 2 sin2x - √ 3 / 2 cos2x = sin2xcos π / 3-sin π / 3cos2x = sin (2X - π / 3) so the minimum positive period of (1) f (x) = 2 π / 2 = π (2) find the monotone increasing interval of F (x)

The solution of noel's equation 2x-5 = 1 is the same as that of 1-third 3a-x = 0, a =?
If you can, here's another question:
When x = - 2, the value of equation (2-m) x + 4 = 18, then when x = 3, the value of equation is?
The solutions of noel's equations 3x-2m + 1 = 0 and 2-m = 2x are opposite to each other, and the value of M is obtained

The solution of the equation 2x-5 = 1 is x = 31 - (3a-x) / 3 = 0, which is also x = 1. Substituting in a = 4 / 3, substituting x = - 2, we get (2-m) (- 2) + 4 = 18, solving M = 9 equation (2-m) x + 4 = - 7x + 4. When x = 3, the value of the equation is - 17, the solution of the equation 3x-2m + 1 = 0 is x = (2m-1) / 3, the solution of the equation 2-m = 2x is x = 1-m / 2, the solutions are opposite, and the sum is 0