Again, please clarify the question. Either "x=y^2/cos(x)*pi " or "x=y^2/cos(pi)". From the question it is not possible to tell whether the second "x" is a variable, or a multiplier sign (and if it were a multiplier, you're question is omitting a variable on the cos(x)).
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The derivative with respect to 'x' of sin(pi x) ispi cos(pi x)
cos pi over four equals the square root of 2 over 2 This value can be found by looking at a unit circle. Cos indicates it is the x value of the point pi/4 which is (square root 2 over 2, square root 2 over 2)
y = 2(x) - (pi/3) + (sqrt(3)/2)
sin x - cos x = 0sin x = cos x(sin x)^2 = (cos x)^2(sin x)^2 = 1 - (sin x)^22(sin x)^2 = 1(sin x)^2 = 1/2sin x = ± √(1/2)sin x = ± (1/√2)sin x = ± (1/√2)(√2/√2)sin x = ± √2/2x = ± pi/4 (± 45 degrees)Any multiple of 2pi can be added to these values and sine (also cosine) is still ± √2/2. Thus all solutions of sin x - cos x = 0 or sin x = cos x are given byx = ± pi/4 ± 2npi, where n is any integer.By choosing any two integers , such as n = 0, n = 1, n = 2 we can find some solutions of sin x - cos x = 0.n = 0, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(0)(pi) = ± pi/4 ± 0 = ± pi/4n = 1, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(1)(pi) = ± pi/4 ± 2pi = ± 9pi/4n = 2, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(2)(pi) = ± pi/4 ± 4pi = ± 17pi/4
cos x - 1 = 0 cos(x) = 1 x = 0 +/- k*pi radians where k = 1,2,3,...