Question:
$\sin x+\cos x=1+\sin x\cos x,$ then
A) $\sin(x+\fracπ4)=\frac1{√2}$
B) $\sin(x-\fracπ4)=\frac1{√2}$
C) $\cos(x+\fracπ4)=\frac1{√2}$
D) $\cos(x-\fracπ4)=\frac1{√2}$
Method $1$:
Let $\sin x+\cos x=t,$ then $1+2\sin x\cos x=t^2$
Therefore, $t=1+\frac{t^2-1}{2}$
$t^2-2t+1=0\implies t=1$
So, $\sin x+\cos x=1$, thus, options A) and D) are correct.
Method $2$:
Putting $\sin x=0$ in the given equation, I get $\cos x=1$, thus, $x=2nπ$. So, options A, C, D are correct.
Putting $\cos x=0$ in the given equation, I get $\sin x=1$, thus, $x=(4n+1)\fracπ2$. So, option B is also correct.
How to streamline the answers of both the methods?
Best Answer
As pointed out in the comments, you cannot assume $\sin x = 0$ or $\cos x = 0$. For instance, $\sin x = 0$ at $x = \pi$, but $\pi$ is not a solution of the equation $\sin x + \cos x = 1 + \sin x\cos x$. Similarly, $\cos x = 0$ at $x = 3\pi/2$, but $x = 3\pi/2$ is also not a solution of the equation $\sin x + \cos x = 1 + \sin x\cos x$.
Here is another approach. \begin{align*} \sin x + \cos x & = 1 + \sin x\cos x\\ 0 & = 1 - \sin x - \cos x + \sin x\cos x\\ 0 & = 1 - \sin x - \cos x(1 - \sin x)\\ 0 & = (1 - \sin x)(1 - \cos x) \end{align*} Since a product is equal to zero if and only if one of its factors is equal to zero, \begin{align*} 1 - \sin x & = 0 & \text{or} & & 1 - \cos x & = 0\\ 1 & = \sin x & & & 1 & = \cos x\\ \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z} & = x & & & 2m\pi, m \in \mathbb{Z} & = x \end{align*} If $x = \frac{\pi}{2} + 2k\pi$, then equations A, B, and D are true. If $x = 2m\pi$, then equations A, C, and D are true. Thus, only equations A and D are true at all solutions of the equation $\sin x + \cos x = 1 + \sin x\cos x$.