Question

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

Answer 1

Hint: We have to find the general solution of the given equation. We will first multiply and divide the left-hand side of the given equation by \[\sqrt 2 \] and will further simplify the equation. As we can compare the obtained equation by \[\sin \left( {a + b} \right)\] and convert the equation in that form and further expand the expression and thus gets the general solution of the equation.

Complete step by step answer:

We have to find the general solution of the given equation \[\sin \theta + \cos \theta = 1\].
Now, we will multiply and divide the left-hand side of the given equation by \[\sqrt 2 \].
Thus, we get,
\[ \Rightarrow \sqrt 2 \left[ {\sin \theta \cdot \dfrac{1}{{\sqrt 2 }} + \cos \theta \cdot \dfrac{1}{{\sqrt 2 }}} \right] = 1\]
As we know that \[\cos 45 = \dfrac{1}{{\sqrt 2 }}\] and \[\sin 45 = \dfrac{1}{{\sqrt 2 }}\]. Thus, we get,
\[ \Rightarrow \sin \theta \cos 45 + \cos \theta \sin 45 = \dfrac{1}{{\sqrt 2 }}\]
As we can compare the left-hand side of the above equation by \[\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\].
Thus, we get,
\[ \Rightarrow \sin \theta \cos 45 + \cos \theta \sin 45 = \sin \left( {\theta + \dfrac{\pi }{4}} \right)\]
We can also write the expression \[\sin \left( {\theta + \dfrac{\pi }{4}} \right)\] using the expansion of \[\sin \left( {\theta + \alpha } \right) = \sin \left( {n\pi + {{\left( { - 1} \right)}^n}\alpha } \right)\] as follows,
\[ \Rightarrow \sin \left( {\theta + \dfrac{\pi }{4}} \right) = \sin \left( {n\pi + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right)\]
On further simplifying we get,
\[
   \Rightarrow \theta + \dfrac{\pi }{4} = n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{4} \\
   \Rightarrow \theta = n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{4} - \dfrac{\pi }{4} \\
   \Rightarrow \theta = n\pi + \left[ {{{\left( { - 1} \right)}^n} - 1} \right]\dfrac{\pi }{4} \\
 \]
Thus, with this we can conclude that the general solution of the given equation is \[\theta = n\pi + \left[ {{{\left( { - 1} \right)}^n} - 1} \right]\dfrac{\pi }{4},n = 0, \pm 1, \pm 2\].
Thus, option C is correct.

Note: To find the general solution of the equation, we have to find the value of \[\theta \]. Remember the basic properties like we have used above also such as \[\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\]. We need to remember the expansion of \[\sin \left( {\theta + \alpha } \right) = \sin \left( {n\pi + {{\left( { - 1} \right)}^n}\alpha } \right)\]. We can take any positive or negative value of \[n\] including 0. In the beginning remember to multiply and divide the left side by \[\sqrt 2 \]. Do not forget the basic trigonometric values.