I confused with trigonometry. $\sin x - \cos x = 1$

Question

I got confused with trigonometry.I think that my way which I tried to solve is ok because I don't understand what I just did wrong.

$\sin x - \cos x = 1$

$\sin x - 1 = \cos x$

$\sin x - 1 = \sqrt{1 - \sin^2(x)} \rightarrow$ Now I think that I can square both sides, but before that I find definitions for $x$ : $f(x) = g(x); g(x) \geq 0$ so $'f(x)'$ also must be greater or equal $0$.

         f(x)>=0 ==> [sinx - 1 >= 0]

                     [sinx >= 1] ==> x=90°

$[1 - \sin^2(x) \geq 0]$

$[\sin^2(x) \leq 1]$

$[\sin^2(x) - 1 \leq 0]$

$[(\sin x - 1)(\sin x + 1) \leq 0]$

$(\sin x - 1 \leq 0) \Rightarrow **x=90°** x \in [0°,180°]$

$(\sin x + 1 \leq 0) ==> **x=270°** x \in [180°,360°]$

https://i.stack.imgur.com/RlwSd.png

So what I have done, it just bring me confusion...

Help me guys with this trigonometry..

Answer 1

$$\frac{1}{\sqrt2}\sin{x}-\frac{1}{\sqrt2}\cos{x}=\frac{1}{\sqrt2}$$ or $$\sin\left(x-45^{\circ}\right)=\sin45^{\circ},$$ which gives $$x-45^{\circ}=45^{\circ}+360^{\circ}k,$$ where $k$ is an integer number, or $$x-45^{\circ}=135^{\circ}+360^{\circ}k,$$ which gives the answer: $$\left\{90^{\circ}+360^{\circ}k,180^{\circ}+360^{\circ}k|,k\in\mathbb Z\right\}.$$

Answer 2

just a hint

$$\sin(x)=1+\cos(x)$$ becomes

$$2\sin(\frac x2)\cos(\frac x2)=2\cos(\frac x2)\cos(\frac x2)$$

thus $$\cos(x/2)=0$$ or $$\sin(x/2)=\cos(x/2)=\sin(\pi/2-x/2)$$

Answer 3

$\sin x -\cos x = 1$

Squaring both sides we get:

$\sin^2 x + \cos^2 x -2 \sin x \cos x=1$

⇒$- \sin 2x= 0 = \sin 4k\pi+\pi$

⇒ $x=(4k+1) \frac{\pi}{2}$; $k ∈ N$

Or $x = (2k+1)\pi $; $k ∈ N$

Answer 4

There's an easy way to solve this problem. Since we have $$ \sin x - \cos x = 1 $$ Then we can square both sides to obtain $$ \sin^2 x + \cos^2 x - 2\sin x \cos x = 1$$ So by the Pythagorean identity means $$ 2\sin x \cos x = 0 $$ Or $$ \sin 2x = 0.$$ From here I think you know what to do.