Question

General solution of
$\sin x+\cos x=\underset{a\in R}{min}\{1,a^2-4a+6\}$ is

• A. $\frac{n\pi }{2}+{(-1)}^n\frac{\pi }{4}$
• B. $2n\pi +{(-1)}^n\frac{\pi }{4}$
• C. $n\pi +{(-1)}^{n+1}\frac{\pi }{4}$
• D. $n\pi +{(-1)}^n\frac{\pi }{4}-\frac{\pi }{4}$

Answer 1

The correct option is D $n\pi +{(-1)}^n\frac{\pi }{4}-\frac{\pi }{4}$

Given that,
$\sin x+\cos x=\underset{a\in R}{min}\{1,a^2-4a+6\}$
Now, $a^2-4a+6={(a-2)}^2+2$
$\therefore \underset{a\in R}{min}\{1,a^2-4a+6\}=min\{1,2\}=1$
$\therefore \sin x+\cos x=1$
$\Rightarrow \sin (x+\frac{\pi }{4})=\frac{1}{\sqrt{2}}$
$\Rightarrow x+\frac{\pi }{4}=n\pi +{(-1)}^n\cdot \frac{\pi }{4}$
$\Rightarrow x=n\pi +{(-1)}^n\frac{\pi }{4}-\frac{\pi }{4}$