Find Number of solutions of $\sin (2x)+\cos (2x)+\sin x+\cos x=1$ in $\left [0 \:\: 2\pi\right]$

The equation can be written as:

$$\sin (2x)+1-2 \sin^2x+\sin x+\cos x=1$$ $\implies$

$$\sin x+\cos x=2\sin^2 x-2 \sin x\cos x$$ $\implies$

$$\sin x+\cos x=2\sin x\left(\sin x-\cos x\right)$$


$$\frac{\sin x+\cos x}{\sin x-\cos x}=2\sin x$$ $\implies$

$$\frac{1+\tan x}{1-\tan x}=-2\sin x$$

$$\tan \left(\frac{\pi}{4}+x\right)=-2\sin x$$

Now i have drawn the graphs of $\tan \left(\frac{\pi}{4}+x\right)$ and $-2\sin x$ and observed there are two solutions.

is there any other way?


Another way would be to let $t=\tan(\frac x2)$ and arrive to $$t^4+2 t^3+8 t^2-6 t-1=0$$ Now, using the formulae for the quartic equation, the discriminant is $\Delta=-309248$ which shows that the equation has two distinct real roots and two complex conjugate non-real roots.


I think your reasoning with graphs is not so right because if so,

why not to draw the graph of $f(x)=\sin2x+\cos2x+\sin{x}+\cos{x}-1$?

By the Claude's hint from your equation $$\frac{1+\tan{x}}{1-\tan{x}}=-2\sin{x}$$ after substitution $\tan\frac{x}{2}=t$ we obtain $$\frac{1+\frac{2t}{1-t^2}}{1-\frac{2t}{1-t^2}}=-2\cdot\frac{2t}{1+t^2}$$ or $$t^4+2t^3+8t^2-6t-1=0.$$

Now, let $$f(t)=t^4+2t^3+8t^2-6t-1.$$ Thus, $$f''(t)=12t^2+12t+16>0,$$ which says that $f$ is a convex function.

Hence, a graph of $f$ and the $t$-axis have two common points maximum.

But $f(0)<0$, which says that $f$ has two roots exactly and since the period of $\tan$ is $\pi$, we get that the starting equation has two roots exactly on $[0,2\pi]$.


  • $\begingroup$ why my reasoning is wrong? its very difficult to draw original expression $\endgroup$ – Ekaveera Kumar Sharma Oct 24 '17 at 9:59
  • $\begingroup$ @Ekaveera Kumar Sharma I said that it's not so right because you can not draw these graphs exactly. By the way, if you can make this then to draw graphs of $y=\sin{x}$, $y=\cos{x}$, $y=\sin2x$ and $y=\cos2x$ and to add, it's also possible, but it's not gives the solution of your problem, I think. $\endgroup$ – Michael Rozenberg Oct 24 '17 at 10:03

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