Solve the following equation: $\sqrt {\sin x - \sqrt {\cos x + \sin x} } = \cos x$

Question

Solve the following equation: \begin{array}{l}{\sqrt{\sin x-\sqrt{\cos x+\sin x}}=\cos x} \\ \text{my try as follows:}\\{\sin x-\sqrt{\cos x+\sin x}=\cos ^{2} x} \\ {\sin x-\cos ^{2} x=\sqrt{\cos x+\sin x}} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x=\cos x+\sin x} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x-\cos x-\sin x=0} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x\left(1-\sin ^{2} x\right)-\cos x-\sin x=0} \\{\sin ^2}x + {\left( {1 - {{\sin }^2}x} \right)^2} - 2\sin x\left( {1 - {{\sin }^2}x} \right) - \cos x - \sin x = 0\\ {\sin ^{2} x+\sin ^{4} x-2 \sin ^{2} x+1-2 \sin x+2 \sin ^{3} x-\cos x-\sin x=0} \\ {\sin ^{4} x+2 \sin ^{3} x-\sin ^{2} x-3\sin x-\cos x+1=0}\end{array}

Now i think it gets more complicated , any help would be appreciated

Answer 1

We do this in one period $[-\pi, \pi)$.

According to the equation, $\cos x = \sqrt \cdots \geqslant 0$, then $x \in [-\pi/2, \pi/2]$. Also inside the square root, $\sin x \geqslant \sqrt \cdots \geqslant 0$, then $x\in [0, \pi) \cup \{-\pi\}$. Thus $x \in [0, \pi/2]$ at least.

To be more exact, we consider $$ \sin x \geqslant \sqrt {\sin x + \cos x}, $$ equivalently, $$ \sin^2 x \geqslant \sin x + \cos x. $$ But in $[0, \pi/2]$, $\cos x \geqslant 0$ and $\sin^2 x \leqslant \sin x$, thus $$ \sin x \geqslant \sin^2 x \geqslant \sin x + \cos x \geqslant \sin x, $$ which means all $\geqslant $ are $=$. The $=$ holds iff $\cos x = 0$ and $\sin x = 1$. Thus the only possible $x$ is $\boldsymbol {\pi/2}$.

Clearly $x = \pi/2$ is truly a solution to the equation, so the final solutions are $$ \boxed {2n\pi +\frac \pi 2, n \in \mathbb Z }\ . $$

Answer 2

We see that $\sin{x}\geq0$ and $\cos{x}\geq0.$

Let $f(x)=\sqrt{\sin{x}-\sqrt{\sin{x}+\cos{x}}},$ where $x\in\left[0,\frac{\pi}{2}\right].$

Thus, $$f'(x)=\frac{\cos{x}-\frac{\cos{x}-\sin{x}}{2\sqrt{\sin{x}+\cos{x}}}}{2\sqrt{\sin{x}-\sqrt{\sin{x}+\cos{x}}}}>0$$ for $\cos{x}-\sin{x}<0.$

But for $\cos{x}-\sin{x}\geq0$ we obtain: $$f'(x)=\frac{4\cos^2x(\sin{x}+\cos{x})-(\cos{x}-\sin{x})^2}{4\sqrt{\sin{x}+\cos{x}}\sqrt{\sin{x}-\sqrt{\sin{x}+\cos{x}}}(2\cos{x}\sqrt{\sin{x}+\cos{x}}+\cos{x}-\sin{x})}$$ and $$4\cos^2x(\sin{x}+\cos{x})-(\cos{x}-\sin{x})^2=4\cos^2x(\sin{x}+\cos{x})+2\sin{x}\cos{x}-1=$$ $$=4\cos^2x(\sin{x}+\cos{x})+2\sin{x}\cos{x}-\sqrt{\sin^2x+\cos^2x}\geq$$ $$\geq2\cos^2x(\sin{x}+\cos{x})-\sin{x}-\cos{x}=(\sin{x}+\cos{x})^2(\cos{x}-\sin{x})\geq0,$$ which says that $f$ increases.

But $\cos$ decreases, which says that there is unique $\alpha\in\left[0,\frac{\pi}{2}\right]$, for which our equation has root.

$\frac{\pi}{2}$ is a root, which gives the answer: $$\left\{\frac{\pi}{2}+2\pi k|k\in\mathbb Z\right\}$$

Answer 3

It can be seen that the radical of the form $\sqrt{x-\sqrt{x+..}}$ is undefined for $x<1$. Since $\sin{x}$ is on the interval $[-1,1]$. This means that it is only defined for $x=\arcsin(1)$. Or generalizing this, we get; $$x=2n\pi+\frac{\pi}{2}, \tag{for n ∈ Z}$$

Answer 4

In any solution, $\cos(x)$ is nonnegative, since it equals some square root.

In any solution, $\sin(x)$ is nonnegative, since the equation includes the square root of $\sin(x)$ minus some nonnegative quantity.

So $\cos(x)+\sin(x)\geq\sin(x)$, and therefore $$\sqrt{\cos(x)+\sin(x)}\geq\sqrt{\sin(x)}\geq\sin(x)$$ where the last inequality is because $\sin(x)\leq1$. And note that equality only holds if $\sin(x)=1$ or $\sin(x)=0$.

So $\sin(x)-\sqrt{\cos(x)+\sin(x)}\leq\sin(x)-\sin(x)=0$, and therefore $$\sin(x)-\sqrt{\cos(x)+\sin(x)}\leq0$$

In the equation, we take a square root of this expression. So it is not defined unless we have equality, which we noted is only possible if $\sin(x)=1$ or $\sin(x)=0$.

No solution can have $\sin(x)=0$, or else $\sin(x)-\sqrt{\cos(x)+\sin(x)}<0$, and we cannot take its square root in the equation.

Any $x$ with $\sin(x)=1$ implies $\cos(x)=0$, and such $x$ do indeed make a solution, by inspection.

So the given equation is equivalent to $\sin(x)=1$, and the solution set is $\frac{\pi}{2}+2k\pi$.