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) $$x=2n\pi +{\displaystyle \frac{1}{2}}\pi$$
(c) $$x=n\pi +{(-1)}^n{\displaystyle \frac{\pi }{4}}-{\displaystyle \frac{\pi }{4}}$$
(d) None of these

Answer 1

Hint: Firstly, consider an expression of the form $$r\sin (x+a)=1$$. Expand this expression using the formula, $$\sin (A+B)=\sin A\cos B+\cos A\sin B$$ 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:
$$r\sin (x+a)=1$$.
$$r\sin \cos a+r\cos x\sin a=1$$.
We have written the above equation using the trigonometry compound angle formula which is given below:
$$\sin (A+B)=\sin A\cos B+\cos A\sin B.$$
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:
$$\begin{array}{rl}& r\cos a=1\\ & \Rightarrow \cos a={\displaystyle \frac{1}{r}}.........(i)\\ & r\sin a=1\\ & \Rightarrow \sin a={\displaystyle \frac{1}{r}}.........(ii)\end{array}$$
Dividing the above-mentioned equations, we get:
$${\displaystyle \frac{\sin a}{\cos a}}={\displaystyle \frac{{\displaystyle \frac{1}{r}}}{{\displaystyle \frac{1}{r}}}}$$
$$\Rightarrow tana=1$$
But we know, $\tan {\displaystyle \frac{\pi }{4}}=1$
So, the value of a is $${\displaystyle \frac{\pi }{4}}$$.
As we know that $${\cos }^{2}a+{\sin }^{2}a=1$$, substituting values from equation (i) and (ii), we can rewrite this equation as:
$${\displaystyle \frac{1}{r^2}}+{\displaystyle \frac{1}{r^2}}=1$$.
$$\Rightarrow {\displaystyle \frac{2}{r^2}}=1$$
$$r^2=2$$
Taking the square root on both sides, we get
$$r=\sqrt{2}$$.
Substituting the value of ‘r’ and ‘a’ in the assumed equation$$r\sin (x+a)=1$$, we have:
$$\sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})=1$$
$$\sin (x+{\displaystyle \frac{\pi }{4}})={\displaystyle \frac{1}{\sqrt{2}}}$$
But we know, $\sin \left({\displaystyle \frac{\pi }{4}}\right)={\displaystyle \frac{1}{\sqrt{2}}}$ , substituting this value the above equation can be written as,
$$\sin (x+{\displaystyle \frac{\pi }{4}})=\sin {\displaystyle \frac{\pi }{4}}$$
The general solution of this equation is
$$\Rightarrow x+{\displaystyle \frac{\pi }{4}}=n\pi +{(-1)}^n{\displaystyle \frac{\pi }{4}}$$, for $$n=0,\pm 1,\pm 2,\pm 3..........$$
Therefore $$x=n\pi +{(-1)}^n{\displaystyle \frac{\pi }{4}}-{\displaystyle \frac{\pi }{4}}$$ where $$n\in z$$, because if $$\sin x=\sin y$$ then x is given as $$x=n\pi +{(-1)}^{n}y$$.
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 $$\sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})=1$$ directly, thus saving time. Using identities in solving trigonometric functions plays a key role in these types of questions.