Question
The general solution of the trigonometric equation $$\sin x + \cos x = 1$$ is given by:
A.
$$x = 2n\pi ;\,\,n = 0,\,\, \pm 1,\,\, \pm 2\,.....$$
B.
$$x = 2n\pi + \frac{\pi }{2};\,\,n = 0,\,\, \pm 1,\,\, \pm 2\,.....$$
C.
$$x = n\pi + {\left( { - 1} \right)^n}\,\,\frac{\pi }{4} - \frac{\pi }{4}$$
D.
none of these
Answer :
$$x = n\pi + {\left( { - 1} \right)^n}\,\,\frac{\pi }{4} - \frac{\pi }{4}$$
Solution :
$$\eqalign{
& \sin x + \cos x = 1 \cr
& \Rightarrow \,\,\frac{1}{{\sqrt 2 }}\sin x + \frac{1}{{\sqrt 2 }}\cos = \frac{1}{{\sqrt 2 }} \cr
& \Rightarrow \,\,\sin x\cos \frac{\pi }{4} + \cos x\sin \frac{\pi }{4} = \sin \frac{\pi }{4} \cr
& \Rightarrow \,\,\sin \left( {x + \frac{\pi }{4}} \right) = \sin \frac{\pi }{4} \cr
& \Rightarrow \,\,x + \frac{\pi }{4} = n\pi + {\left( { - 1} \right)^n}\frac{\pi }{4},n \in Z\left( {{\text{the set of integers}}} \right) \cr
& \Rightarrow \,\,x = n\pi + {\left( { - 1} \right)^n}\frac{\pi }{4} - \frac{\pi }{4} \cr
& {\text{where }}n = 0, \pm 1, \pm 2,..... \cr} $$