Question

The general solution of the equation \[\sin x + \cos x = 1\] is
A. \[x = 2n\pi + \dfrac{\pi }{2},n = 0, \pm 1, \pm 2\]
B. \[x = n\pi + \left( {{{\left( { - 1} \right)}^n} + 1} \right)\dfrac{\pi }{4},n = 0, \pm 1, \pm 2\]
C. \[x = n\pi + \left( {{{\left( { - 1} \right)}^n} - 1} \right)\dfrac{\pi }{4},n = 0, \pm 1, \pm 2\]
D. \[x = 2n\pi ,n = 0, \pm 1, \pm 2\]

Answer 1

- Hint: First of all, divide both sides with \[\sqrt {{{\left( {{\text{Coefficient os }}\sin x} \right)}^2} + {{\left( {{\text{Coefficient of }}\cos x} \right)}^2}} \]. Then split the terms to make the whole equation in terms of sine angles by using the formula \[\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)\] to obtain the required answer.


Complete step-by-step solution -

Given equation is \[\sin x + \cos x = 1\]
Dividing both sides with \[\sqrt {{{\left( {{\text{Coefficient os }}\sin x} \right)}^2} + {{\left( {{\text{Coefficient of }}\cos x} \right)}^2}} \], we get
\[
  \dfrac{1}{{\sqrt {{1^2} + {1^2}} }}\left[ {\sin x + \cos x} \right] = \dfrac{1}{{\sqrt {{1^2} + {1^2}} }} \\
  \dfrac{{\sin x + \cos x}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }} \\
\]
Splitting the terms, we have
\[\dfrac{{\sin x}}{{\sqrt 2 }} + \dfrac{{\cos x}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}\]
As\[\cos {45^0} = \sin {45^0} = \dfrac{1}{{\sqrt 2 }}\], we have
\[\sin x\cos {45^0} + \cos x\sin {45^0} = \dfrac{1}{{\sqrt 2 }}\]
We know that \[\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)\]
\[\sin \left( {x + {{45}^0}} \right) = \dfrac{1}{{\sqrt 2 }}\]
We know that, the general solution of \[\sin x = \dfrac{1}{{\sqrt 2 }}\] is \[x = n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{4}\]
\[\sin \left( {x + \dfrac{\pi }{4}} \right) = \sin \left( {n\pi + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right)\]
Cancelling sine terms on both sides, we get
\[
  x + \dfrac{\pi }{4} = n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{4} \\
  x = n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{4} - \dfrac{\pi }{4} \\
  x = n\pi + \left( {{{\left( { - 1} \right)}^n} - 1} \right)\dfrac{\pi }{4}{\text{ where }}n{\text{ is a integer}} \\
\]
Thus, the correct option is C. \[x = n\pi + \left( {{{\left( { - 1} \right)}^n} - 1} \right)\dfrac{\pi }{4},n = 0, \pm 1, \pm 2\]

 Note: The general solution of \[\sin x = \dfrac{1}{{\sqrt 2 }}\] is \[x = n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{4}\]. If you didn’t see the option of your obtained solution, then convert the terms into cosine angles by using the formula \[\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B\].