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

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

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

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 $$p(t)=(3t^2-1)(t^2+2t+3)$$.
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 $$t=-1$$.