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The number of solutions of the equation `sin^3xcosx+sin^2xcos^2x+sinxcos^3x=1` in the interval `[0,2pi]` is/are 0 (b) 2 (c) 3 (d) infinite

A

0

B

2

C

3

D

infinite

Text Solution

Verified by Experts

The correct Answer is:
A

`sin^(2) x cos x+sin^(2) x cos^(2) x+ sin x cos^(3)x=1`
or `sin x cos x (sin^(2) x+sin x cos x + cos^(2)x)=1`
or `(sin 2x)/2 (1+(sin 2x)/2)=1`
or `sin 2x(2+sin 2x)=4`
or `sin^(2) 2x+2 sin 2x-4 =0`
or `sin 2x=(-2 pm sqrt(4+16))/2=-1 pm sqrt(5)`
This is not possible since `-1 le sin 2x le 1`.
Hence, the given equation has no solution.
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