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Question: The most general values of x for which \[\sin x + \cos x = {\min _{a \in \mathbb{R}}}\\{ 1,{a^2} - 4...

The most general values of x for which sin⁡x+cos⁡x=min⁡a∈R1,a2−4a+6\sin x + \cos x = {\min _{a \in \mathbb{R}}}\\{ 1,{a^2} - 4a + 6\\} are given by
a. 2nπ;n∈Z2n\pi ;n \in \mathbb{Z}
b. 2nπ+π2;n∈Z2n\pi + \dfrac{\pi }{2};n \in \mathbb{Z}
c. nπ+(−1)n.π4−π4;n∈Zn\pi + {( - 1)^n}.\dfrac{\pi }{4} - \dfrac{\pi }{4};n \in \mathbb{Z}
d. 2nπ−π2;n∈Z2n\pi - \dfrac{\pi }{2};n \in \mathbb{Z}

Explanation

Solution

To start with we can use the fact that what is the minimum value of the given polynomial, a2−4a+6{a^2} - 4a + 6. If the value is less than 1, then we can proceed by using that value for sin⁡x+cos⁡x\sin x + \cos x to find the general value for x.

Complete step-by-step answer:
Given sin⁡x+cos⁡x=min⁡a∈R1,a2−4a+6\sin x + \cos x = {\min _{a \in \mathbb{R}}}\\{ 1,{a^2} - 4a + 6\\}
We have now,
a2−4a+6{a^2} - 4a + 6
On splitting last term, we get
=a2−4a+4+2= {a^2} - 4a + 4 + 2
Using a2−2ab+b2=(a−b)2{a^2} - 2ab + {b^2} = {(a - b)^2} , we get,
=(a−2)2+2= {(a - 2)^2} + 2
So, minimum, a2−4a+6{a^2} - 4a + 6
=(0)2+2= {(0)^2} + 2
On simplification we get,
= 0 + 2$$$$ = 2
So, now, sin⁡x+cos⁡x=min⁡(1,a2−4a+6)\sin x + \cos x = \min (1,{a^2} - 4a + 6)
So, the minimum value is, sin⁡x+cos⁡x=1\sin x + \cos x = 1
Then, sin⁡x+cos⁡x=1\sin x + \cos x = 1
Multiplying by 12\dfrac{1}{{\sqrt 2 }} , we get,
⇒12sin⁡x+12cos⁡x=12\Rightarrow \dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x = \dfrac{1}{{\sqrt 2 }}
Using, cos⁡π4=12\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}and sin⁡π4=12\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} , we get,
⇒sin⁡π4sin⁡x+cos⁡π4cos⁡x=cos⁡π4\Rightarrow \sin \dfrac{\pi }{4}\sin x + \cos \dfrac{\pi }{4}\cos x = \cos \dfrac{\pi }{4}
Now using cos⁡(a−b)=cos⁡acos⁡b+sin⁡asin⁡b\cos (a - b) = \cos a\cos b + \sin a\sin b , we get,
⇒cos⁡(x−π4)=cos⁡π4\Rightarrow \cos (x - \dfrac{\pi }{4}) = \cos \dfrac{\pi }{4}
Using, x=2nπ±yx = 2n\pi \pm y for cos⁡x=cos⁡y\cos x = \cos y, we get,
⇒x−π4=2nπ±π4\Rightarrow x - \dfrac{\pi }{4} = 2n\pi \pm \dfrac{\pi }{4}
On simplification we get,
⇒x=2nπ±π4+π4\Rightarrow x = 2n\pi \pm \dfrac{\pi }{4} + \dfrac{\pi }{4}
Then, x=2nπ+π2x = 2n\pi + \dfrac{\pi }{2} or x=2nπx = 2n\pi
Also, As, sin⁡x+cos⁡x=1\sin x + \cos x = 1
Multiplying by 12\dfrac{1}{{\sqrt 2 }} , we get,
⇒12sin⁡x+12cos⁡x=12\Rightarrow \dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x = \dfrac{1}{{\sqrt 2 }}
Using, cos⁡π4=12\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}and sin⁡π4=12\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}, we get,
⇒cos⁡π4sin⁡x+sin⁡π4cos⁡x=sin⁡π4\Rightarrow \cos \dfrac{\pi }{4}\sin x + \sin \dfrac{\pi }{4}\cos x = \sin \dfrac{\pi }{4}
Using sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b\sin (a + b) = \sin a\cos b + \cos a\sin b , we get,
⇒sin⁡(x+π4)=sin⁡π4\Rightarrow \sin (x + \dfrac{\pi }{4}) = \sin \dfrac{\pi }{4}
Using, x=nπ+(−1)nyx = n\pi + {( - 1)^n}yfor sin⁡x=sin⁡y\sin x = \sin y
⇒x+π4=nπ+(−1)nπ4\Rightarrow x + \dfrac{\pi }{4} = n\pi + {( - 1)^n}\dfrac{\pi }{4}
On simplification we get,
⇒x=nπ+(−1)nπ4−π4\Rightarrow x = n\pi + {( - 1)^n}\dfrac{\pi }{4} - \dfrac{\pi }{4} where n∈Zn \in \mathbb{Z}
We have the general value as, x=nπ+(−1)nπ4−π4x = n\pi + {( - 1)^n}\dfrac{\pi }{4} - \dfrac{\pi }{4} where n∈Zn \in \mathbb{Z} which is option c.

Note: We have the general value of sin⁡x=sin⁡y\sin x = \sin yas, x=nπ+(−1)nyx = n\pi + {( - 1)^n}y and also cos⁡x=cos⁡y\cos x = \cos y as, x=2nπ±yx = 2n\pi \pm y . Then we get the values of x as, the value of n goes on starting from, n=1,2,3n = 1,2,3,……. And so on.