The answer depends on x,
on whether x is measured in degrees or radians
and on whether the question is about sin(x) + 2 or sin(x + 2).
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F(x) = 6 sin(x) + 2 cos(x)F'(x) = 6 cos(x) - 2 sin(x)
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
The differentiation of sin x plus cosx is cos (x)-sin(x).
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
2*sin^2(x) - 5*sin(x) + 2 = 0 is a quadratic equation in sin(x).therefore,{2*sin(x) - 1}*{sin(x) - 2)} = 0=> sin(x) = 1/2 or sin(x) = 2The second solution is rejected since sin(x) cannot exceed 1.The principal solution is x = arcsin(1/2) = pi/6 radians. Additional or alternative solutions will depend on the domain for x - which has not been given.