1.

The number of roots of the equation `sin(2x+pi/18) cos(2x-pi/9)=-1/4` in `[0, 2pi]` isA. 2B. 4C. 6D. 8

Answer» Correct Answer - B
`2sin(2x+(pi)/(18)).cos(2x-(2pi)/(18))=(-1)/(2)`
`therefore 2 sin (2x + 10^(@))cos (2x-20^(@))=(-1)/(2)`
`therefore sin (4x-10^(@))+sin 30^(@)=(-1)/(2)`
`therefore sin (4x-10^(@))=-1=sin 270^(@)`
Now period of `sin (4x - 10^(@))` is `pi//2`
For `x in (0, pi//2)`
`therefore 4x = 280^(@)`
`therefore x = 70^(@)`
`therefore` for x in `[0, 2pi]`, there are four solutions


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