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Solve `2 sin^(3) x=cos x`. |
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Answer» Correct Answer - `x= npi +pi/4, n in Z` Since `sin x=0` does not satisfy the equation, dividing the equation by `sin^(3)x` we get `cot x osec^(2) x=2` or `cot^(3)x+cot x-2=0` or `(cot x-1) (cot^(2) x+cot x+2)=0` or `cot x=1` `rArr x=npi+pi//4, n in Z` |
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