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Real solution of equation $$(\cos x -\sin x)\cdot \bigg(2\tan x+\frac{1}{\cos x}\bigg)+2=0.$$

Try: Using Half angle formula

$\displaystyle \cos x=\frac{1-\tan^2x/2}{1+\tan^2 x/2}$ and $\displaystyle \sin x=\frac{2\tan^2 x/2}{1+\tan^2 x/2}$

Substuting These values in equation

we have an polynomial equation in terms of $t=\tan x/2$

So our equation $$3t^{4}+6t^{3}+8t^{2}-2t-3=0$$ Could Some Help me how to Factorise it.

OR is there is any easiest way How to solve it, Thanks

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    $\begingroup$ $\displaystyle \sin x=\frac{2\tan x/2}{1+\tan^2 x/2}$. Also, $x=\pi$ is a real solution of the equation, but you can't get that from your substitution. $\endgroup$ – Shubham Johri Nov 27 '18 at 7:59
  • $\begingroup$ Are you sure $x=\pi$ is a solution? $\endgroup$ – user Nov 27 '18 at 8:10
  • $\begingroup$ Whoops, my bad. $\endgroup$ – Shubham Johri Nov 27 '18 at 8:33
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That’s a nice way to solve but recall that $\sin x=\frac{2t}{1+t^2}$.

I didn’t check whether it is only a typo or a mistake in the derivation but note that wolfy suggests $$p(t)=(3t^2-1)(t^2+2t+3)$$.

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HINT:

Since $\sin x=\frac{2t}{1+t^2}$, the equation becomes $$\left(\frac{1-t^2-2t}{1+t^2}\right)\left(\frac{2t+1+t^2}{1-t^2}\right)+2=0\implies -t^4-4t^3-4t^2+1=2t^4-2$$ or $$3t^4+4t^3+4t^2-3=0$$ from which a root is $t=-1$.

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  • $\begingroup$ It is very far from a hint in that way and I think the asker can get it by his/herself. $\endgroup$ – user Nov 27 '18 at 8:13

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