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in JEE by (251 points)

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1 Answer

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As per above question their is basic form used:

sin(2x) - √3 sin(x) 
2sin(x) cos(x) - √3 sin(x) 
Sin(x) (2cos(x) - √3)

To find its root let's equate it to zero
 Sin(x) (2cos(x) - √3) = 0
Either, Sin(x) = 0 or cos(x) = √3 / 2
CAlculate its root.
Now the graph is

For area integrate the curve from a to b with its values obtained respectively.
Also it is periodic function in x with period 2π

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