[" 14."quad " zभीड "Sigma" e "" 6"" s "(sqrt(3)x)+cos^(-1)x=(pi)/(2)],[" - "(1)/(2)]

cos^(-1)x sqrt(3)+cos^(-1)x=(pi)/(2)

Solve : e^(sin x+sqrt(3) cosx-1)=1 (-2pi lt x lt 2pi)

Prove that : "cos"^(-1)sqrt((2)/(3))-"cos"^(-1)(sqrt(6)+1)/(2sqrt(3))=(pi)/(6)

Prove that (i) "cos " ((pi)/(3) +x) =(1)/(2) ( " cos " x - sqrt(3) sin x) (ii) " sin " ((pi)/(4) + x) + " sin " ((pi)/(4)-x) =sqrt(2) " cos " x (iii) (1)/(sqrt(2)) " cos ((pi)/(4) + x) = (1)/(2) " (cos x - sin x) " (iv) " cos x + cos " ((2pi)/(3) +x) + " cos " ((2pi)/(3)-x) =0

int_(0)^( pi/6)(sqrt(3cos2x-1))/(cos x)dx

lim_ (x rarr (pi) / (6)) (2-sqrt (3) cos x-sin x) / ((6x-pi) ^ (2))

The number of solutions for the equation 2 sin^(-1)(sqrt(x^(2) - x + 1)) + cos^(-1)(sqrt(x^(2) - x) )= (3pi)/(2) is

The wave function of 3s and 3p_(z) orbitals are given by : Psi_(3s) = 1/(9sqrt3) ((1)/(4pi))^(1//2) ((Z)/(sigma_(0)))^(3//2)(6=6sigma+sigma)e^(-sigma//2) Psi_(3s_(z))=1/(9sqrt6)((3)/(4pi))^(1//2)((Z)/(sigma_(0)))^(3//2)(4-sigma)sigmae^(-sigma//2)cos0, sigma=(2Zr)/(nalpha_(0)) where alpha_(0)=1st Bohr radius , Z= charge number of nucleus, r= distance from nucleus. From this we can conclude:

The wave function of 3s and 3p_(z) orbitals are given by : Psi_(3s) = 1/(9sqrt3) ((1)/(4pi))^(1//2) ((Z)/(sigma_(0)))^(3//2)(6-6sigma+sigma)e^(-sigma//2) Psi_(3p_(z))=1/(9sqrt6)((3)/(4pi))^(1//2)((Z)/(sigma_(0)))^(3//2)(4-sigma)sigmae^(-sigma//2)cos0, sigma=(2Zr)/(nalpha_(0)) where alpha_(0)=1st Bohr radius , Z= charge number of nucleus, r= distance from nucleus. From this we can conclude:

The number of the solutions of the equation 2 sin^(-1) sqrt(x^(2) + x + 1) + cos^(-1) sqrt(x^(2) + x) = (3pi)/(2) is