1.

Find all the values of x,in between the range [0,2pi] for 2sin(3x)+3cos(3x)=0

Answer»

LET y^2 = X^2 - 3x

y^2 - y - 2 = 0

y^2 - 2y + y - 2 = 0

y(y - 2) + 1 (y - 2) = 0

(y+1)(y-2) = 0

y = -1 or y = 2

Then

By resubstitution -

x^2 - 3x = -1

x^2 - 3x + 1 = 0

By shreedharacharya formula -

x = {3 ± √(9–4)}/4

x = ( 3 ± √5 ) / 4

Also -

x^2 - 3x = 2

x^2 - 3x - 2 = 0

By shreedharacharya formula -

x = {3 ± √(9+16)}/4

x = ( 3 ± 5 ) / 4

x = 2 or x = -1/2

x = 2, -1/2, (3+√5)/4, (3-√5)/4

A2A. THANK-YOU!


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