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

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.