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ત્રિકોણમિતીય સમીકરણો અને ત્રિકોણના ગુણધર્મો

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ગણિત ધોરણ 11 સેમિસ્ટર 2

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નીચેનું સમીકરણ ઉકેલો :
2 space cos space straight theta space plus space sec space straight theta space equals space 3


therefore space 2 space cos space straight theta space plus space fraction numerator 1 over denominator cos space straight theta space end fraction space equals space 3

therefore space 2 space cos squared space straight theta space minus space 3 space cos space straight theta space plus space 1 space equals space 0

therefore space 2 space cos squared space straight theta space minus space 2 space cos space straight theta space minus space cos space space straight theta space plus space 1 space equals space 0

therefore space open parentheses 2 space cos space straight theta space minus space 1 close parentheses space open parentheses cos space straight theta space minus space 1 close parentheses space equals space 0

therefore space 2 space cos space straight theta space minus space 1 space equals space 0 space અથવ ા space cos space straight theta space minus space 1 space equals space 0

therefore space cos space straight theta space equals space 1 half space અથવ ા space cos space straight theta space equals space 1

therefore space cos space straight theta space equals space 1 half space equals space cos space straight pi over 3 space અથવ ા space cos space straight theta space equals space 1 space equals space cos space 0

therefore space straight theta space equals space 2 kπ space plus-or-minus space straight pi over 3 space comma space straight k space element of space straight Z space અથવ ા space straight theta space equals space 2 kπ space plus-or-minus space 0 space comma space straight k space element of space straight Z

therefore space straight theta space equals space 2 kπ space comma space straight k space element of space straight Z space અથવ ા space straight theta space equals space 2 kπ space plus-or-minus space straight pi over 3 space comma space straight k space element of space straight Z

therefore space ઉક ે લગણ space colon space open curly brackets bold 2 bold kπ bold space left enclose bold space bold k bold space bold element of bold space bold Z end enclose close curly brackets bold space bold union bold space open curly brackets bold 2 bold kπ bold space bold plus-or-minus bold space bold pi over bold 3 bold space left enclose bold space bold k bold space bold element of bold space bold Z end enclose close curly brackets

નીચેનું સમીકરણ ઉકેલો :
square root of 2 space cos e c space 3 theta space minus space 2 space equals space 0


therefore space cosec space 3 straight theta space equals space square root of 2

therefore space sin space 3 straight theta space equals space fraction numerator 1 over denominator square root of 2 end fraction

therefore space sin space 3 straight theta space equals space fraction numerator 1 over denominator square root of 2 end fraction space equals space sin space straight pi over 4

therefore space 3 straight theta space equals space kπ space plus space open parentheses negative 1 close parentheses to the power of straight k space straight pi over 4 comma space straight k space element of space straight Z

therefore space straight theta space equals space kπ over 3 space plus space open parentheses negative 1 close parentheses to the power of straight k space straight pi over 12 space comma space straight k space element of space straight Z

therefore space ઉક ે લગણ space colon bold space open curly brackets bold kπ over bold 3 bold space bold plus bold space open parentheses bold minus bold 1 close parentheses to the power of bold k bold space bold pi over bold 12 bold space left enclose bold space bold k bold space bold element of bold space bold Z end enclose close curly brackets

નીચેનું સમીકરણ ઉકેલો :
2 space cos space 2 straight theta space plus space square root of 2 space equals space 0


therefore space cos space 2 straight theta space equals space minus fraction numerator 1 over denominator square root of 2 end fraction space equals space cos space open parentheses straight pi space minus space straight pi over 4 close parentheses space equals space cos space open parentheses fraction numerator 3 straight pi over denominator 4 end fraction close parentheses

therefore space cos space 2 straight theta space equals space cos space open parentheses fraction numerator 3 straight pi over denominator 4 end fraction close parentheses

therefore space 2 straight theta space equals space 2 kπ space plus-or-minus space fraction numerator 3 straight pi over denominator 4 end fraction comma space straight k space element of space straight Z

therefore space straight theta space equals space kπ space plus-or-minus space fraction numerator 3 straight pi over denominator 8 end fraction space comma space straight k space element of space straight Z

therefore space ઉક ે લગણ space colon space open curly brackets kπ space plus-or-minus space fraction numerator 3 straight pi over denominator 8 end fraction space left enclose space straight k space element of space straight z end enclose close curly brackets

નીચેનું સમીકરણ ઉકેલો :
2 space cos squared space straight theta space plus space square root of 3 space cos space straight theta space equals space 0


therefore space cos space straight theta space open parentheses 2 space cos space straight theta space plus space square root of 3 close parentheses space equals space 0

therefore space cos space straight theta space equals space 0 space અથવ ા space cos space straight theta space equals space minus fraction numerator square root of 3 over denominator 2 end fraction space equals space cos space open parentheses straight pi space minus space straight pi over 6 close parentheses space equals space cos space open parentheses fraction numerator 5 straight pi over denominator 6 end fraction close parentheses

therefore space straight theta space equals space open parentheses 2 straight k space plus space 1 close parentheses straight pi over 2 comma space straight k space element of space straight Z space અથવ ા space cos space straight theta space equals space cos space open parentheses fraction numerator 5 straight pi over denominator 6 end fraction close parentheses

therefore space straight theta space equals space open parentheses 2 straight k space plus space 1 close parentheses space straight pi over 2 comma space straight k space element of space straight Z space અથવ ા space straight theta space equals space 2 kπ space plus-or-minus space fraction numerator 5 straight pi over denominator 6 end fraction comma space straight k space element of space straight Z

therefore space ઉક ે લગણ space colon space open curly brackets open parentheses bold 2 bold k bold space bold plus bold space bold 1 close parentheses bold pi over bold 2 bold space left enclose bold space bold k bold space bold element of bold space bold Z end enclose close curly brackets bold space bold union bold space open curly brackets bold 2 bold kπ bold space bold plus-or-minus bold space fraction numerator bold 5 bold pi over denominator bold 6 end fraction bold space left enclose bold space bold k bold space bold element of bold space bold Z end enclose close curly brackets

નીચેનું સમીકરણ ઉકેલો :
4 space sin squared space straight theta space minus space 8 space cos space straight theta space plus space 1 space equals space 0


therefore space 4 space minus space 4 space cos squared space straight theta space minus space 8 space cos space straight theta space plus space 1 space equals space 0

therefore space 4 space cos squared space straight theta space plus space 8 space cos space straight theta space minus space 5 space equals space 0

therefore space 4 space cos squared space straight theta space plus space 10 space cos space straight theta space minus space 2 space cos space straight theta space minus space 5 space equals space 0

therefore space 2 space cos space straight theta space open parentheses 2 space cos space straight theta space plus space 5 close parentheses space minus space 1 space open parentheses 2 space cos space straight theta space plus space 5 close parentheses space equals space 0

therefore space open parentheses 2 space cos space straight theta space minus space 1 close parentheses space open parentheses 2 space cos space straight theta space plus space 5 close parentheses space equals space 0

therefore space cos space straight theta space equals space 1 half space અથવ ા space cos space straight theta space equals space minus 5 over 2

પરંતુ cos space theta space equals space minus 5 over 2 space less than space minus 1 , જે શક્ય નથી.

હવે, cos space theta space equals space 1 half

therefore space cos space straight theta space equals space 1 half space equals space cos space straight pi over 3

therefore space straight theta space equals space 2 kπ space plus-or-minus space straight pi over 3 space comma space straight k space element of space straight Z

therefore space ઉક ે લગણ space colon bold space open curly brackets bold 2 bold kπ bold space bold plus-or-minus bold space bold pi over bold 3 bold space left enclose bold space bold k bold space bold element of bold space bold Z end enclose close curly brackets