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Home / Doubts and Solutions/JEE/Mathematics

kindly solve (ref. 230)

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Asked by arup.isro 11th July 2018, 11:38 PM
Answered by Expert
Answer:
begin mathsize 16px style 2 sin squared straight theta space plus space sin squared open parentheses 2 straight theta close parentheses equals 2 2 sin squared straight theta space plus open parentheses 2 space sinθ space cos space straight theta close parentheses squared equals 2 2 sin squared straight theta space plus 4 space sin squared straight theta space cos squared straight theta equals 2 sin squared straight theta space plus 2 space sin squared straight theta space cos squared straight theta equals 1 sin squared straight theta space plus 2 space sin squared straight theta space open parentheses 1 minus sin squared straight theta close parentheses equals 1 2 sin to the power of 4 straight theta minus 3 sin squared straight theta plus 2 equals 0 put space sin squared straight theta equals straight x 2 straight x squared minus 3 straight x plus 2 equals 0 straight x equals fraction numerator 3 plus-or-minus square root of 9 minus 8 end root over denominator 2 end fraction sin squared straight theta equals 1 space or fraction numerator 1 over denominator square root of 2 end fraction sin space straight theta space equals plus-or-minus 1 straight theta space equals space straight pi over 2 plus 2 πn space space or space straight theta space equals space minus straight pi over 2 plus 2 πn sin space straight theta space equals plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction straight theta space equals space straight pi over 4 plus πn space space or space straight theta space equals space minus straight pi over 4 plus πn end style
Answered by Expert 12th July 2018, 11:08 AM
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