Number of real solution of complex Trigonometric equation

Question

Number of real solution of the equation $$\sin^5 x-\cos^5 x=\frac{1}{\cos x}-\frac{1}{\sin x}.(\cos x \neq \sin x)$$

Try: $$\;\; \sin^4 x+\sin^3 x\cdot \cos x+\sin^2 x\cdot \cos^2 x+\sin x\cdot \cos^3 x+\cos^4 x=\frac{1}{\sin x\cos x}$$

Now divide both side by $\cos^6 x\neq 0$

$$\tan^5 x+\tan^4 x+\tan^3 x+\tan^2 x+\tan x=(1+\tan^2 x)^3$$

Put $\tan x = t\;,$ Then $t^5+t^4+t^3+t^2+t+1=(1+t^2)^3$

So $$t^5+t^4+t^3+t^2+t=t^6+1+3t^4+3t^2$$

$$t^6-t^5+2t^4-t^3+2t^2-t+1=0$$

Now i am struck here, Could some help me how i find real $t$ thanks

Answer 1

Hint: Your last equation can be factorized as $$(t^2+t+1)(t^2-t+1)^2=0$$

Answer 2

Here is another method. Using $s=\sin x, c=\cos x$ you get to:

$$s^4+s^3c+s^2c^2+sc^3+c^4=\frac 1{sc}$$ then you can try to reduce the degree by using $s^2+c^2=1$ so you have that $$s^3c+sc^3=sc(s^2+c^2)=sc$$ and $$s^4+s^2c^2+c^4=(s^2+c^2)^2-s^2c^2=1-s^2c^2$$

Now let $sc=u$ and you get $$1+u-u^2=\frac 1u$$ or $$u^3-u^2-u+1=0$$which factors as $$(u-1)(u^2-1)=(u-1)^2(u+1)=0$$

I saw the factors because the coefficients were symmetric, so the reciprocal of any solution would be a solution. There would be at least one real solution for $u$.

Now note that $u=\frac 12\sin 2x$