Menu Close

7cos-2-x-sin-2-x-3-2cos-2-x-sin-2-x-czm-7cos-2-x-1-cos-2-x-3-2cos-2-x-sin-2-x-6cos-2-x-2-2cos-2-x-1-cos-2-x-6cos-2-x-2-3cos-2-x-1-2-3cos-2-x-1-3cos-2-x-1-2-




Question Number 11901 by ahmet last updated on 04/Apr/17
((7cos^2 x+sin^2 x−3)/(2cos^2 x−sin^2 x))=?  czm∵  ((7cos^2 x+1−cos^2 x−3)/(2cos^2 x−sin^2 x))  ((6cos^2 x−2)/(2cos^2 x−(1−cos^2 x)))=((6cos^2 x−2)/(3cos^2 x−1))  ((2(3cos^2 x−1))/(3cos^2 x−1))=2
$$\frac{\mathrm{7}{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}−\mathrm{3}}{\mathrm{2}{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}}=? \\ $$$${czm}\because\:\:\frac{\mathrm{7}{cos}^{\mathrm{2}} {x}+\mathrm{1}−{cos}^{\mathrm{2}} {x}−\mathrm{3}}{\mathrm{2}{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}} \\ $$$$\frac{\mathrm{6}{cos}^{\mathrm{2}} {x}−\mathrm{2}}{\mathrm{2}{cos}^{\mathrm{2}} {x}−\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)}=\frac{\mathrm{6}{cos}^{\mathrm{2}} {x}−\mathrm{2}}{\mathrm{3}{cos}^{\mathrm{2}} {x}−\mathrm{1}} \\ $$$$\frac{\mathrm{2}\left(\mathrm{3}{cos}^{\mathrm{2}} {x}−\mathrm{1}\right)}{\mathrm{3}{cos}^{\mathrm{2}} {x}−\mathrm{1}}=\mathrm{2} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *