Trigonometrical Equations Ques 10

  1. 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: 10.(c)

Solution:

Formula:

Trigonometric Equations:

  1. 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} $



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