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ત્રિકોણમિતીય સમીકરણો અને ત્રિકોણના ગુણધર્મો

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ગણિત ધોરણ 11 સેમિસ્ટર 2

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ગણિત

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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

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


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


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


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


નીચેનું સમીકરણ ઉકેલો :
sin space 2 theta space plus space cos space theta space equals space 0


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