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# 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 Verified
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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:
\begin{align} & r\cos a=1 \\ & \Rightarrow \cos a=\dfrac{1}{r}.........(i) \\ & r\sin a=1 \\ & \Rightarrow \sin a=\dfrac{1}{r}.........(ii) \\ \end{align}
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.
Last updated date: 23rd Sep 2023
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