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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.imgur.com/wBHZ2MC.png

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

Help me guys with this trigonometry..

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  • $\begingroup$ Claiming that $\cos x= \sqrt{1-\sin^2x}$ is wrong. A byproduct of this is that, when you apply the correct procedure of discussing the sign of $A$ in $A=\sqrt B$, you should end up losing part of the solutions. $\endgroup$
    – user562983
    Jun 30, 2018 at 21:56
  • $\begingroup$ How? If $sin^2x + cos^2x = 1 -> cosx = √(1- sin^2x)$ $\endgroup$
    – Hury H
    Jun 30, 2018 at 21:58
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    $\begingroup$ Since when does $A^2=B$ imply $A=\sqrt B$? That's never been a thing. $\endgroup$
    – user562983
    Jun 30, 2018 at 21:59
  • $\begingroup$ 1. Follow some of the links that show you how to use MathJax better. Your formulas are still not very easy to read. 2. Explain in words what you did in each step. Just writing a formula is usually not enough. 3. What is the point of the image you linked to? $\endgroup$
    – David K
    Jun 30, 2018 at 22:13
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    $\begingroup$ Please read this tutorial on how to typeset mathematics on this site using MathJax. $\endgroup$
    – N. F. Taussig
    Jun 30, 2018 at 22:22

4 Answers 4

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

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$$\frac{1}{\sqrt2}\sin{x}-\frac{1}{\sqrt2}\cos{x}=\frac{1}{\sqrt2}$$ 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: $$\left\{90^{\circ}+360^{\circ}k,180^{\circ}+360^{\circ}k|,k\in\mathbb Z\right\}.$$

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$\sin x -\cos x = 1$

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 $; $k ∈ N$

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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.

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