40*cos^2(x)=18*sin(x)+31

40*cos^2(x)=18*sin(x)+31 im Intervall 23pi-25pi

$40\times cos^{2}x=18\times sin\ x+31$

$cos^2x=1-sin^2x$

$40\times(1-sin^2x)=18\times sin\ x+31$      [s = sin x

$40- 40s^2=18s+31$

$40s^2+18s -9=0$

  a           b          c

$s = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$s = {-18 \pm \sqrt{18^2-4\times 40\times (-9)} \over 2\times 40}$

$sin\ x=\frac{-18\pm\sqrt{1764}}{80}$

$sin\ x_1=0,3$

$x_1=arcsin\ 0,3=0,30469 (+24\pi\ 25\pi \ 26\pi)$

$sin\ x_2=-0,75$

$x_2=arcsin(-0,75)=-0,84806(+24\pi\ 25\pi\ 26\pi)$

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