Question

The number of solutions of the equation sin³x cosx+ sin²x cos²x+sinxcos³x =1 in the interval [0, 2π] is

• A. 0
• B. 2
• C. 3
• D. Infinite

Answer 1

Answer: (A)

0

Answer 2

Here sin³x cosx + sin²x cos² x + sinx cos³ x =1 ⇒ sinx

cosx ( sin²x + sinx cosx+ cos² x)=1 ⇒ $\frac { \sin 2 x } { 2 } \left( 1 + \frac { \sin 2 x } { 2 } \right)$

= 1 ⇒ sin2x(2+ sin2x) = 4

⇒ sin²2x + 2 sin2x -4 = 0

⇒ sin2x = $\frac { - 2 \pm \sqrt { 4 + 16 } } { 2 }$ ⇒ sin2x = -1 ± $\sqrt { 5 }$

This is not possible .

( since –1 ≤ sin2x ≤ 1)

Hence the given equation has no solution.