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58. sin 3x + sin 2x – sin x = 4sin x cosx2  cos3x2

0 2 Views | Posted 4 months ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    58. L.H.S = sin 3x + sin 2x - sin x

    = sin 3x - sin x + sin 2x.

    = 2 cos 3x+x2 . sin 3x−x2 + sin 2x [?sinA−sinB=2cos  A+B2  sin  A−B2]

    = 2 cos 4x2 sin 2x2 + sin 2x.

    = 2 cos 2x sin x + 2 sin xcosx        [ ? sin 2x = 2sin xcosx]

    = 2 sin x [cos 2x + cosx]

    = 2 sin x [2cos  2x+x2  cos  2x−x2] [? cosA+cos  B=2cos  A+B2  cos  A−B2]

    = 2 sin x [2.cos3x2cosx2]

    = 4 sin xcos x2 . cos 3x2 = R.H.S.

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Option (4) is correct.

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12x =

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one root lies in ( − 1 2 ,   0 ) , one root is c o s π 1 2  which is positive. As the coefficients are real, therefore all the roots must be real.

Statement 2 is false.

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