Answer
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Hint: In this question we have to find the value of $x$ so the key concept is to simplify the given equation $\sin (\pi x) + \cos \left( {\pi x} \right) = 0$ by using some basic trigonometric formulas.
“Complete step-by-step answer:”
We have been given that $\sin (\pi x) + \cos \left( {\pi x} \right) = 0$ ………... (1)
Now let’s simplify the equation (1)
$
\Rightarrow \sin (\pi x) + \cos (\pi x) = 0 \\
\Rightarrow \sin (\pi x) = - \cos (\pi x) \\
\Rightarrow \dfrac{{\sin (\pi x)}}{{\cos (\pi x)}} = - 1 \\
\\
$ ………….. (2)
And we know that $\dfrac{{\sin x}}{{\cos x}} = \tan x$
So the equation (2) can be written as
$
\Rightarrow \dfrac{{\sin (\pi x)}}{{\cos \left( {\pi x} \right)}} = - 1 \\
\Rightarrow \tan \left( {\pi x} \right) = - 1 \\
\\
$ …………. (3)
Since domain $\tan \theta $ is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$
Therefore at $\theta = - \dfrac{\pi }{4}$
$ \Rightarrow \tan ( - \dfrac{\pi }{4}) = - \tan \dfrac{\pi }{4} = - 1$ ………….. (4)
Now in equation (3) replace $ - 1$ by $\tan \left( { - \dfrac{\pi }{4}} \right)$ we get,
$
\Rightarrow \tan (\pi x) = \tan ( - \dfrac{\pi }{4}) \\
\Rightarrow \pi x = - \dfrac{\pi }{4} \\
\Rightarrow x = - \dfrac{1}{4} \\
$
Now to recheck the solution putting the value of $x = - \dfrac{1}{4}$ in L.H.S of equation (1) we get,
$
\Rightarrow \sin ( - \dfrac{\pi }{4}) + \cos ( - \dfrac{\pi }{4}) \\
\Rightarrow - \sin \dfrac{\pi }{4} + \cos \dfrac{\pi }{4} \\
\Rightarrow - \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} = 0 = {\text{R}}{\text{.H}}{\text{.S}} \\
$
Hence at $x = - \dfrac{1}{4}$ equation (1) is satisfied.
And hence the value of $x = - \dfrac{1}{4}$ for which $\sin (\pi x) + \cos (\pi x) = 0$.
Note: Whenever we face such types of problems the key point is that first we find the domain of the given equation and then find the value corresponding to that domain only and equating to it. After getting the value, we should recheck the given solution by putting in the given equation whether the solution satisfies the given equation or not. If the solution satisfies the given equation it means that the founded solution was our right answer.
“Complete step-by-step answer:”
We have been given that $\sin (\pi x) + \cos \left( {\pi x} \right) = 0$ ………... (1)
Now let’s simplify the equation (1)
$
\Rightarrow \sin (\pi x) + \cos (\pi x) = 0 \\
\Rightarrow \sin (\pi x) = - \cos (\pi x) \\
\Rightarrow \dfrac{{\sin (\pi x)}}{{\cos (\pi x)}} = - 1 \\
\\
$ ………….. (2)
And we know that $\dfrac{{\sin x}}{{\cos x}} = \tan x$
So the equation (2) can be written as
$
\Rightarrow \dfrac{{\sin (\pi x)}}{{\cos \left( {\pi x} \right)}} = - 1 \\
\Rightarrow \tan \left( {\pi x} \right) = - 1 \\
\\
$ …………. (3)
Since domain $\tan \theta $ is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$
Therefore at $\theta = - \dfrac{\pi }{4}$
$ \Rightarrow \tan ( - \dfrac{\pi }{4}) = - \tan \dfrac{\pi }{4} = - 1$ ………….. (4)
Now in equation (3) replace $ - 1$ by $\tan \left( { - \dfrac{\pi }{4}} \right)$ we get,
$
\Rightarrow \tan (\pi x) = \tan ( - \dfrac{\pi }{4}) \\
\Rightarrow \pi x = - \dfrac{\pi }{4} \\
\Rightarrow x = - \dfrac{1}{4} \\
$
Now to recheck the solution putting the value of $x = - \dfrac{1}{4}$ in L.H.S of equation (1) we get,
$
\Rightarrow \sin ( - \dfrac{\pi }{4}) + \cos ( - \dfrac{\pi }{4}) \\
\Rightarrow - \sin \dfrac{\pi }{4} + \cos \dfrac{\pi }{4} \\
\Rightarrow - \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} = 0 = {\text{R}}{\text{.H}}{\text{.S}} \\
$
Hence at $x = - \dfrac{1}{4}$ equation (1) is satisfied.
And hence the value of $x = - \dfrac{1}{4}$ for which $\sin (\pi x) + \cos (\pi x) = 0$.
Note: Whenever we face such types of problems the key point is that first we find the domain of the given equation and then find the value corresponding to that domain only and equating to it. After getting the value, we should recheck the given solution by putting in the given equation whether the solution satisfies the given equation or not. If the solution satisfies the given equation it means that the founded solution was our right answer.
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