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Solve the Following Equation: Sin X + Cos X = 1 - Mathematics

Sum

Solve the following equation:

\[\sin x + \cos x = 1\]
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Solution

Given:
\[\sin x + \cos x = 1\]      ...(i)

The equation is of the form 
\[a \sin \theta + b \cos \theta = c\], where 
\[a = 1, b = 1\] and \[c = 1\].
Let: \[a = r \sin \alpha\] and \[b = r \cos \alpha\]
Now,
\[r = \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{2}\] and

\[\tan\alpha = \frac{b}{a} = 1 \Rightarrow \alpha = \frac{\pi}{4}\]

On putting

\[a = 1 = r \sin \alpha\] and \[b = 1 = r \cos \alpha\] in equation (i), we get:
\[r \sin \alpha \sin x + r \cos \alpha \cos x = 1\]

\[\Rightarrow r \cos ( x - \alpha) = 1\]

\[ \Rightarrow \sqrt{2} \cos \left( x - \frac{\pi}{4} \right) = 1\]

\[ \Rightarrow \cos \left( x - \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}\]

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

\[ \Rightarrow x - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4}, n \in Z\]

On taking positive sign, we get:
\[x - \frac{\pi}{4} = 2n\pi + \frac{\pi}{4}\]
\[ \Rightarrow x = 2n\pi + \frac{\pi}{4} + \frac{\pi}{4}\]
\[ \Rightarrow x = 2n\pi + \frac{\pi}{2}, n \in Z\]
On taking negative sign, we get:
\[x - \frac{\pi}{4} = 2m\pi - \frac{\pi}{4}\]
\[ \Rightarrow x = 2m\pi, m \in Z\]

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 11 Trigonometric equations
Exercise 11.1 | Q 6.3 | Page 22
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