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Question: Find the value of \(x\) for which \(\sin (\pi x) + \cos \left( {\pi x} \right) = 0\)....

Find the value of xx for which sin⁡(πx)+cos⁡(πx)=0\sin (\pi x) + \cos \left( {\pi x} \right) = 0.

Explanation

Solution

Hint: In this question we have to find the value of xx so the key concept is to simplify the given equation sin⁡(πx)+cos⁡(πx)=0\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⁡(πx)+cos⁡(πx)=0\sin (\pi x) + \cos \left( {\pi x} \right) = 0 ………... (1)
Now let’s simplify the equation (1)
⇒sin⁡(πx)+cos⁡(πx)=0 ⇒sin⁡(πx)=−cos⁡(πx) ⇒sin⁡(πx)cos⁡(πx)=−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 sin⁡xcos⁡x=tan⁡x\dfrac{{\sin x}}{{\cos x}} = \tan x
So the equation (2) can be written as
⇒sin⁡(πx)cos⁡(πx)=−1 ⇒tan⁡(πx)=−1   \Rightarrow \dfrac{{\sin (\pi x)}}{{\cos \left( {\pi x} \right)}} = - 1 \\\ \Rightarrow \tan \left( {\pi x} \right) = - 1 \\\ \\\ …………. (3)
Since domain tan⁡θ\tan \theta is (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)
Therefore at θ=−π4\theta = - \dfrac{\pi }{4}
⇒tan⁡(−π4)=−tan⁡π4=−1\Rightarrow \tan ( - \dfrac{\pi }{4}) = - \tan \dfrac{\pi }{4} = - 1 ………….. (4)
Now in equation (3) replace −1 - 1 by tan⁡(−π4)\tan \left( { - \dfrac{\pi }{4}} \right) we get,
⇒tan⁡(πx)=tan⁡(−π4) ⇒πx=−π4 ⇒x=−14  \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=−14x = - \dfrac{1}{4} in L.H.S of equation (1) we get,
⇒sin⁡(−π4)+cos⁡(−π4) ⇒−sin⁡π4+cos⁡π4 ⇒−12+12=0=R.H.S  \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=−14x = - \dfrac{1}{4} equation (1) is satisfied.
And hence the value of x=−14x = - \dfrac{1}{4} for which sin⁡(πx)+cos⁡(πx)=0\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.