Question
If x sin3 θ + y cos3 θ = sin θ
cos θ and x sin θ − y cos θ = 0, then the value of x2 + y2 equalsSolution
x sin3 θ + y cos3 θ = sin θ cos θ -> eq 1 x sin θ − y cos θ = 0 x sin θ = y cos θ -->eq2 substituting in eq1 y cos θ sin2 θ + y cos3 θ = sin θ cos θ taking y cos θ common y cos θ(sin2 θ + cos2 θ) = sin θ cos θ { we know sin2 θ + cos2 θ =1} y cos θ = sin θ cos θ y = sin θ substituting in eq 2 x sin θ = sin θ cos θ x = cos θ x2 + y2 sin2 θ + cos2 θ = 1
I. 8x² - 78x + 169 = 0
II. 20y² - 117y + 169 = 0
(i) 2x² – x – 3 = 0
(ii) 2y² – 6y + 4 = 0
I.√(3x-17)+ x=15
II. y + 135/y=24
I. 5x + 2y = 31
II. 3x + 7y = 36
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 50x + 600 = 0
Equation 2: y² - 51y + 630 = 0
I. 15y2 + 26y + 8 = 0
II. 20x2 + 7x – 6 = 0
I. 9/(4 )p + 7/8p = 21/12
II. 7/5p = 9/10q + 1/4
- Suppose both the roots of q² + kq + 49 = 0 are real and equal, then determine the value of 'k'.
I. 2x² - 9x + 10 = 0
II. 3y² + 11y + 6 = 0
l). p² - 29p + 204 = 0
ll). q² + 4q - 221 = 0