I have been trying to solve $\sin x - \cos x = 1$ by squaring both sides but has not been able to obtain the solution. Here is what I did:
$$\begin{align}(\sin x - \cos x)^2 &=1^2\\\sin^2x-2\sin x\cos x+\cos^2x&=1\\1-\sin 2x&=1\\\sin2x&=0\end{align}$$
Obviously $x=0$ is not a solution. May I ask why this is the case or where did things go wrong?
Thank You
Use the property that $$\sin(x-y)=\sin x\cos y-\cos x\sin y$$
We can rewrite the left side of the equation....
$$\begin{align}\sin x-\cos x&=1\\\sqrt 2\left(\sin x\cos\dfrac\pi4-\cos x\sin\dfrac\pi4\right)&=1\\\sin\left(x-\dfrac\pi4\right)&=\dfrac1{\sqrt2}\\x-\dfrac\pi4&=\dfrac\pi4+2k\pi,\pi-\dfrac\pi4+2k\pi\\x&=\boxed{\dfrac\pi2+2k\pi,(2k+1)\pi}\end{align}$$
Don't square both sides, it creates extraneous solutions
If you want, you can also use this property to start the solution....
$$\begin{align}a\cos x+b\sin x&\equiv R\cos(x-\alpha)\\\text{where }R=\sqrt{a^2+b^2},&\qquad\alpha=\tan^{-1}\left(\dfrac ba\right)\end{align}$$
You can not use squaring because $a=b$ is not equal to $a^2=b^2$.
After squaring you obtain also solutions of the following equations. $$\sin{x}-\cos{x}=-1,$$ which we don't want.
I think it's better the following way. $$2\sin\frac{x}{2}\cos\frac{x}{2}-2\cos^2\frac{x}{2}=0$$ or $$\cos\frac{x}{2}(\sin\frac{x}{2}-\cos\frac{x}{2})=0,$$ which gives $x=\pi+2\pi k$, $k\in\mathbb Z$ or $$\tan\frac{x}{2}=1,$$ which gives $x=\frac{\pi}{2}+2\pi k$.
Note that : $$\sin x - \cos x = \sqrt2 \sin(x - \frac{\pi}{4})$$ In this case, $$\sin x - \cos x = \sqrt2 \sin(x-\frac{\pi}{4})=1 \implies \sin(x-\frac{\pi}{4})=\frac{1}{\sqrt2}$$
Thus, two possible solutions are:
$$x - \frac{\pi}{4}=\frac{\pi}{4} \implies x = \frac{\pi}{2}+2k\pi \, ,k \in \mathbb{Z}$$
$$x -\frac{\pi}{4}=\frac{3\pi}{4} \implies x = \pi + 2k\pi \, , k \in \mathbb{Z}$$
I hope you know the expansion$$\sin(\theta+\varphi)=\sin\theta\cos\varphi+\sin\varphi\cos\theta$$Because we can use it in the equation and greatly simplify the algebra needed. Rewriting your equation$$\sin x-\cos x=1\quad\implies\quad\sqrt2\left(\frac {\sin x}{\sqrt2}-\frac {\cos x}{\sqrt2}\right)=1$$The expression inside the parenthesis can be rewritten as$$\sin x\cos\frac {\pi}4-\cos x\sin\frac {\pi}4=\sin\left(x-\frac {\pi}4\right)$$Hence$$\sqrt2\sin\left(x-\frac {\pi}4\right)=1$$Can you take it from here?
After squaring both sides of the equarion, we get $$ 1=\sin^2x+\cos^2x-2\sin x\cos x=1-\sin2x. $$ Hence $$ \sin 2x=0 $$ i.e. $$ x=x_k=\frac{\pi}{2}k, \qquad k \in \mathbb{Z}. $$ Checking with the original equation, we have:
If $k=2n$, then $$ \sin(x_{2n})-\cos(x_{2n})=(-1)^n $$ If $k=2n+1$, then $$ \sin(x_{2n+1})-\cos(x_{2n+1})=(-1)^n $$ Hence, the solutions are $$ x_{4n}=2\pi n, \quad x_{4n+1}=\frac{\pi}{2}+2\pi n, \qquad n \in \mathbb{Z}. $$
Let $a:= \sin x -\cos x;$ $ b=1;$
You have $a = b.$
Squaring: $a^2 =b^2$ gives
$(a-b)(a+b)=0. $
You find solutions for the squared equation:
$a=b$, or $a = -b.$
Need to be a bit careful:
$\sin 2x =0;$ or $x=0;$ implies:
1) $x = 2πk$, $k$ integer.
These are the solutions for $a = -b.$, I.e.
$\sin x - \cos x = -1$.
The solution to the original problem,
$\sin x - \cos x= +1:$
2) $2x = π$ ; or $x = π/2; $ or
$x = π/2 + 2πk.$
As squaring will yield more answers than enough, I am now going to solve the equation using $\sin ^{2} x+\cos ^{2} x=1$.
Let $a=\sin x \text { and } b=\cos x,$ then
$$ \begin{aligned} &\left\{\begin{array}{l} a-b=1 \quad \cdots(1) \\ a^{2}+b^{2}=1 \quad \cdots(2) \end{array}\right.\\ \Rightarrow \quad & a^{2}+(a-1)^{2}=1 \\ \Rightarrow \quad & 2 a(a-1)=0 \\ \Rightarrow \quad & a=0 \text { or } 1 \end{aligned} $$
A. When $a=0, b=-1$ $$ \left\{\begin{array}{l} \sin x=0 \\ \cos x=-1 \end{array} \Rightarrow x=(2 n+1) \pi\right. \text {, where }n\in Z. $$ B. When $a=1, b=0$ $$ \left\{\begin{array}{l} \sin x=1 \\ \cos x=0 \end{array} \quad \Rightarrow x=\frac{(4 n+1) \pi}{2} \text {, where } n \in \mathbb{Z}\right. $$ Therefore the solutions are $x=(2 n+1) \pi \text{ and } \dfrac{(4 n+1) \pi}{2}, \text {where } n\in Z.$
$$ \begin{aligned} \sin x-\cos x &=1 \\ -\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x &=-\frac{1}{\sqrt{2}} \\ \cos \left(x+\frac{\pi}{4}\right) &=-\frac{1}{\sqrt{2}} \\ x+\frac{\pi}{4} &=2 n \pi \pm \frac{3 \pi}{4} \\ x &=\frac{\pi}{2}(4 n+1) \text { or }(2 n-1) \pi, \text{ where }n\in Z. \end{aligned} $$