Question

Solve the following equation:
\[\sin x + \cos x = \sqrt{2}\]

Answer 1

Given:
\[\sin x + \cos x = \sqrt{2}\] ...(i)
The equation is of the form
\[a \sin x + b \cos x = c\], where 

\[a = 1, b = 1\] and

c = `sqrt2`

Let:

\[a = r \sin \alpha\] and

\[a = r \sin \alpha\]

Now,

\[r = \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{2}\] and

\[r = \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{2}\]

On putting

\[a = 1 = r \sin \alpha\] and

b =1 = r cos α in equation (i), we get:

\[r \sin \alpha \sin x + r \cos \alpha \cos x = \sqrt{2}\]

\[\Rightarrow r \cos (x - \alpha) = \sqrt{2}\]

\[ \Rightarrow \sqrt{2} \cos \left( x - \frac{\pi}{4} \right) = \sqrt{2}\]

\[ \Rightarrow \cos \left( x - \frac{\pi}{4} \right) = 1\]

\[ \Rightarrow \cos \left( x - \frac{\pi}{4} \right) = \cos 0\]

\[ \Rightarrow x - \frac{\pi}{4} = n\pi \pm 0, n \in Z\]

\[ \Rightarrow x = n\pi + \frac{\pi}{4}, n \in Z\]

\[ \Rightarrow x = (8n + 1)\frac{\pi}{4}, n \in Z\]