Trigonometrical Equations 1 Question 17
17. The general solution of the trigonometric equation $\sin x+\cos x=1$ is given by
(1981, 2M)
(a) $x=2 n \pi ; n=0, \pm 1, \pm 2, \ldots$
(b) $x=2 n \pi+\pi / 2 ; n=0, \pm 1, \pm 2, \ldots$
(c) $x=n \pi+(-1)^{n} \frac{\pi}{4}-\frac{\pi}{4} ; n=0, \pm 1, \pm 2, \ldots$
(d) None of the above
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Answer:
Correct Answer: 17. (c)
Solution:
- Given, $\sin x+\cos x=1$
On dividing and multiplying each terms by $\sqrt{2}$, we get
$$ \begin{array}{lc} & \frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x=\frac{1}{\sqrt{2}} \\ \Rightarrow & \sin x \cos \frac{\pi}{4}=\cos x \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \\ \Rightarrow & \sin x+\frac{\pi}{4}=\sin \frac{\pi}{4} \\ \Rightarrow & x+\frac{\pi}{4}=n \pi+(-1)^{n} \frac{\pi}{4} \\ \Rightarrow & x=n \pi+(-1)^{n} \frac{\pi}{4}-\frac{\pi}{4}, n \in I \end{array} $$
