Question

Solve the following equations for x and y

x cos3 y + 3x cos y sin2 y = 14 …… (1)

x sin3 y + 3x cos2 y sin y = 13 …… (2)

Answer 1

Answer:

Hope it will help you.....

Step-by-step explanation:

Case I:

If x ≠ 0 Divide (1) by (2)

(cos ^3 y + 3 cos y.sin^2 y)/(sin^3 y + 3 cos^2 y.sin y) = 14/13 Applying componendo and dividendo, we get

((cos y + sin y) / (cos y - sin y))3 = (14 + 13)/(14 - 13) = 27 = 33

⇒ (cos y + sin y) / (cos y - sin y) or (1 + tan y)/(1 - tan y) = 3/1

Again using componendo and Dividendo we get

2/2 tan y = 4/(2)

4 tan y = 2

⇒ tan y = 1/2 [This is possible in 1st & 3rd quadrant]

y = tan^-1 1/2

In 1 st quadrant,

sin y = 1/√5, cos y = 2/√5

Hence, putting value of y in equation (1)

x [8/(5√5) + 3.2/√5 1/5] = 14 x = 5√5

When y is in 3rd quadrant, sin y = -1/√5 and cos y = -2/√5

Putting value of above in equation (1)

Hence, the value of x is x [-8/(5√5) + 3(-2/√5) 1/√5] = 14

⇒ x = – 5√5

Hence, y = tan^–1 1/2, x = 5√5

 

Answer 2

Answer:

man theta con theta theta

Step-by-step explanation:

hence proved