So, here are two functions $x_1(t)=\cos(2t)-1 \text{ and } x_2(t)=\sin^2(t), I = R$. $C_1x_1(t)+C_2x_2(t)=0$

Here I am stuck, because I try to find the values of $t$ such that I can express $C_1 \text{ or } C_2 $ and show that they must be equal to zero. But in this case, no matter the $t$ I choose I get all expression equal to 0, so I cannot conclude that $C_1 \text{ and} C_2 \text{ must be equal to } 0 \text{ so that the equation } C_1x_1(t)+C_2x_2(t)=0 \text{ is valid}.$ I think may be I can use Wronskian determinant as it tells whether two functions are linearly dependant. But the determinant was very complicated, so I think there must be some other solution.

  • $\begingroup$ you can use this $ \cos(2x)=2\cos^2(x)-1$ And also $\cos^2(x)+ \sin^2(x)=1$ You need to find the value of the constants $(C_1;C_2)$ not the value of $t$. $\endgroup$ – Aryadeva Nov 11 '19 at 20:53
  • $\begingroup$ @Isham Yes, I understand, but we first try to guess the value of $t$ so that either $C_1x_1 \text{ or } C_2x_2 \text{ equals to } 0.$ I don't get how it changes something. Still, if I take $t=0$ Then $x_1=0$ and $x_2=0$ and so with any $t$, ( I think). I don't know how to express $C_2 $, for example, because, if I try to eliminate $C_1x_1$ I also eliminate $C_2x_2$. $\endgroup$ – user Nov 11 '19 at 21:13
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    $\begingroup$ It's normal because $x_1=-2\sin^2(t)=-2x_2$ $\endgroup$ – Aryadeva Nov 11 '19 at 22:41
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    $\begingroup$ @Isham Thank you very much! $\endgroup$ – user Nov 11 '19 at 22:45

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