Question

The general solution of the trigonometric equation $\sin x+\cos x=1$, for $n=0,\pm 1,\pm 2,\pm 3..........$ is given by:
(a) $x=2n\pi$
(b) $x=2n\pi +\dfrac{1}{2}\pi$
(c) $x=n\pi +{{(-1)}^{n}}\dfrac{\pi }{4}-\dfrac{\pi }{4}$
(d) None of these

Answer 1

Hint: Firstly, consider an expression of the form $r\sin \left( x+a \right)=1$. Expand this expression using the formula, $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ and compare with the given trigonometric function in the question. Upon using the required trigonometric identity of sine function and the general solution, we can compute the answer.

Given, $\sin x+\cos x=1$, which is a trigonometric equation.
This is a problem based on finding out the general solution of a trigonometric equation.
To initiate the process, let us assume a trigonometric equation:
$r\sin \left( x+a \right)=1$.
$r\sin \cos a+r\cos x\sin a=1$.
We have written the above equation using the trigonometry compound angle formula which is given below:
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B.$
Now let us compare both the equations:
$r\sin \cos a+r\cos x\sin a=1$with $\sin x+\cos x=1$.
Through that we have:

& r\cos a=1 \\\ & \Rightarrow \cos a=\dfrac{1}{r}.........(i) \\\ & r\sin a=1 \\\ & \Rightarrow \sin a=\dfrac{1}{r}.........(ii) \\\ \end{aligned}$$ Dividing the above-mentioned equations, we get: $$\dfrac{\sin a}{\cos a}=\dfrac{\dfrac{1}{r}}{\dfrac{1}{r}}$$ $$\Rightarrow tana=1$$ But we know, $\tan \dfrac{\pi }{4}=1$ So, the value of a is $$\dfrac{\pi }{4}$$. As we know that $${{\cos }^{2}}a+{{\sin }^{2}}a=1$$, substituting values from equation (i) and (ii), we can rewrite this equation as: $$\dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{r}^{2}}}=1$$. $$\Rightarrow \dfrac{2}{{{r}^{2}}}=1$$ $${{r}^{2}}=2$$ Taking the square root on both sides, we get $$r=\sqrt{2}$$. Substituting the value of ‘r’ and ‘a’ in the assumed equation$$r\sin \left( x+a \right)=1$$, we have: $$\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)=1$$ $$\sin \left( x+\dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$$ But we know, $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ , substituting this value the above equation can be written as, $$\sin \left( x+\dfrac{\pi }{4} \right)=\sin \dfrac{\pi }{4}$$ The general solution of this equation is $$\Rightarrow x+\dfrac{\pi }{4}=n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$$, for $$n=0,\pm 1,\pm 2,\pm 3..........$$ Therefore $$x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}-\dfrac{\pi }{4}$$ where $$n\in z$$, because if $$\sin x=\sin y$$ then x is given as $$x=n\pi +{{\left( -1 \right)}^{n}}y$$. Hence, the correct answer is option (c). Note: Alternatively, you can directly multiply and divide the given trigonometric equation simultaneously with $$\sqrt{2}$$ and achieve $$\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)=1$$ directly, thus saving time. Using identities in solving trigonometric functions plays a key role in these types of questions.