For x∈(0,π), the equation sin x+2 sin 2x -sin 3x =3 has
A
infinitely many solutions
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B
three solutions
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C
one solution
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D
no solution
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Solution
The correct option is D no solution
sin x+2 sin 2x-sin 3x =3 ⇒sinx+4sinxcosx−3sinx+4sin3x=3⇒sinx[−2+4cosx+4(1−cos2x)]=3⇒sinx[2−(4cos2x−4cosx+1)+1]=3⇒3−(2cosx−1)2=3cosecx
Now R.H.S. ≥3
But L.H.S. < 3
Hence, no solution.