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Solve `sin x-3 sin 2x + sin 3x = cos x -3 cos 2x + cos 3x`. |
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Answer» The given equation is `sin x-3 sin 2x + sin 3x=cos x-3 cos 2x+cos 3x` `rArr 2 sin 2x cos x-3 sin 2x=2 cos 2x cos x-3 cos 2x` `rArr sin 2x(2 cos x-3)=cos 2x(2 cos x-3)` `rArr sin 2x=cos 2x" "("As "cos x ne 3//2)` `rArr tan 2x=1` `rArr 2x=n pi+pi//4` `rArr x=(npi)/2+pi/8, n in Z` |
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