Trigonometric equations.

Question

$\sin x + \cos x = \min\{1, k^2-4k+6, k \in \mathbb{R}\}$

I assume that the question looks for the minimum of the two.

Thus, $\sin x + \cos x = \min(1,2)= 1$

$\implies \sin\Big(\dfrac x 2 \Big)\cos\Big(\dfrac x 2\Big)= \sin^2 \Big(\dfrac x2\Big) $. On solving we get, $x = 2n\pi$ or $x= 2n\pi + \dfrac\pi2$ , $n \in \mathbb{I}$.

However, answer given is $x = n\pi + (-1)^n\dfrac \pi 4 - \dfrac \pi 4$.

Where have I gone wrong?

Answer 1

Method$\#1:$,

For $x = n\pi + (-1)^n\dfrac \pi 4 - \dfrac \pi 4$

if $n$ is even, $=2m$(say) where $m$ is any integer $$x=2m\pi$$

What if $n$ is odd, $=2m+1$(say) where $m$ is any integer ?

So, the answer is a compact form of what you have reached at.

Method$\#2:$,

$$\sin x+\cos x=1\iff\sin\left(x+\dfrac\pi4\right)=\sin\dfrac\pi4$$

See general solution here.

Method$\#3:$,

$$\sin x+\cos x=1\iff\cos\left(x-\dfrac\pi4\right)=\cos\dfrac\pi4$$

Answer 2

Since $ k^2-4k+6 =(k-2)^2+2$, then $\min\{1, k^2-4k+6, k \in \mathbb{R}\} = \min\{1,2\} = 1$.

So you have to solve $\sin x + \cos x = 1$

Since $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt 2}$, then

\begin{align} \sin x \sin \frac{\pi}{4} + \cos x \cos \frac{\pi}{4} &= \frac{1}{\sqrt 2} \\ \cos \left(x - \frac{\pi}{4} \right)&= \frac{1}{\sqrt 2} \\ x - \frac{\pi}{4} &= 2\pi k\pm \frac{\pi}{4} \\ x &\in \{2\pi k : k \in \mathbb Z\} \cup \{2\pi k + \frac{\pi}{2} : k \in \mathbb Z\} \end{align}