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 $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$.