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Question: Find the general solution of the question, \(\sin \pi x + \cos \pi x = 0.\) Also find all the soluti...

Find the general solution of the question, sinπx+cosπx=0.\sin \pi x + \cos \pi x = 0. Also find all the solutions in [0, 100].

Explanation

Solution

- According to given in the question we have to solve the given trigonometric expression sinπx+cosπx=0.\sin \pi x + \cos \pi x = 0. in between [0, 100] so, first of all we will try to make the given trigonometric expression in form of sinAcosB+cosAsinA\sin A\cos B + \cos A\sin A so that we can apply the formula as given below:
sin(A+B)=sinAcosB+cosAsinA...................(1)\sin (A + B) = \sin A\cos B + \cos A\sin A...................(1)
So, to convert the given trigonometric expression sinπx+cosπx=0\sin \pi x + \cos \pi x = 0 in form of sinAcosB+cosAsinA\sin A\cos B + \cos A\sin A we have to multiply and divide with 2\sqrt 2 in the numerator and denominator of the expression and then we have to convert 2\sqrt 2 into sin\sin and cos\cos with the help of the formula given below:
sinπ4=2...................(2)\sin \dfrac{\pi }{4} = \sqrt 2 ...................(2)
cosπ4=2..................(3)\cos \dfrac{\pi }{4} = \sqrt 2 ..................(3)
So, now with the help of the (2) and (3) we can apply the formula (1) and after this we have to solve the obtained trigonometric expression where the value of π\pi will be eliminated from the both sides of the expression. After that we can solve the expression for the given range [0, 100]

Complete step-by-step answer:
Step 1: First of all we have to multiply and divide with 2\sqrt 2 in the both sides of the given trigonometric expression to obtain the expression in form of sinAcosB+cosAsinA\sin A\cos B + \cos A\sin A
Hence, on multiplying and divide with 2\sqrt 2 in the numerator and denominator of the given expression,

2(sinπx+cosπx)2=0 2sinπx+2cosπx=0  \Rightarrow \dfrac{{\sqrt 2 (\sin \pi x + \cos \pi x)}}{{\sqrt 2 }} = 0 \\\ \Rightarrow \sqrt 2 \sin \pi x + \sqrt 2 \cos \pi x = 0 \\\

Step 2: Now, we have to use the formula (1) and (2) as mentioned in the solution hint in the obtained expression as obtained in step 1.
sinπ4cosπx+cosπ4sinπx=0\Rightarrow \sin \dfrac{\pi }{4}\cos \pi x + \cos \dfrac{\pi }{4}\sin \pi x = 0
Step 3: Now, we have to convert the obtained trigonometric expression in the form of sin(A+B)\sin (A + B) with the help of the formula (1) as mentioned in the solution hint.
sin(π4+πx)=0\Rightarrow \sin \left( {\dfrac{\pi }{4} + \pi x} \right) = 0
Step 4: On solving the expression obtained in step 3.
(π4+πx)=nπ\Rightarrow \left( {\dfrac{\pi }{4} + \pi x} \right) = n\pi
Where, nZn \in Z
πx=nππ4 πx=π(n14)  \Rightarrow \pi x = n\pi - \dfrac{\pi }{4} \\\ \Rightarrow \pi x = \pi \left( {n - \dfrac{1}{4}} \right) \\\
Now, eliminating π\pi from the both sides of the obtained expression just above.
x=(n14)\Rightarrow x = \left( {n - \dfrac{1}{4}} \right)
Step 5: Now, as given in the question we have to find the solution for the given range [0, 100]
Hence,
n(1,2,3,.........98,100)\Rightarrow n \in (1,2,3,.........98,100)
On substituting the value of n we can obtain the value of x hence,
x(34,74,114,...........,3994)x \in \left( {\dfrac{3}{4},\dfrac{7}{4},\dfrac{{11}}{4},...........,\dfrac{{399}}{4}} \right)

Hence with the help of the formula (1), (2) and (3) we have obtained the value of the given trigonometric expression sinπx+cosπx=0\sin \pi x + \cos \pi x = 0 is x=(n14)x = \left( {n - \dfrac{1}{4}} \right) and range from [0, 100] is x(34,74,114,...........,3994)x \in \left( {\dfrac{3}{4},\dfrac{7}{4},\dfrac{{11}}{4},...........,\dfrac{{399}}{4}} \right)

Note: To obtain the range from [0, 100] as given in the question it is necessary to find the value of x so that we can substitute the value of n(1,2,3,............,99,100)n \in (1,2,3,............,99,100)
To make the given trigonometric expression in form of sinAcosB+cosAsinA\sin A\cos B + \cos A\sin A it is necessary to multiply with 2\sqrt 2 in the both sides of the given trigonometric expression.