Question

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

Answer 1

Hint: - According to given in the question we have to solve the given trigonometric expression $\sin \pi x + \cos \pi x = 0.$ in between [0, 100] so, first of all we will try to make the given trigonometric expression in form of $\sin A\cos B + \cos A\sin A$ so that we can apply the formula as given below:
$\sin (A + B) = \sin A\cos B + \cos A\sin A...................(1)$
So, to convert the given trigonometric expression $\sin \pi x + \cos \pi x = 0$ in form of $\sin A\cos B + \cos A\sin A$ we have to multiply and divide with $\sqrt 2 $ in the numerator and denominator of the expression and then we have to convert $\sqrt 2 $ into $\sin $ and $\cos $ with the help of the formula given below:
$\sin \dfrac{\pi }{4} = \sqrt 2 ...................(2)$
$\cos \dfrac{\pi }{4} = \sqrt 2 ..................(3)$
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 $\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 $\sqrt 2 $ in the both sides of the given trigonometric expression to obtain the expression in form of $\sin A\cos B + \cos A\sin A$
Hence, on multiplying and divide with $\sqrt 2 $ in the numerator and denominator of the given expression,
\[
   \Rightarrow \dfrac{{\sqrt 2 (\sin \pi x + \cos \pi x)}}{{\sqrt 2 }} = 0 \\
   \Rightarrow \sqrt 2 \sin \pi x + \sqrt 2 \cos \pi x = 0 \\
 \]
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.
$ \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 $\sin (A + B)$ with the help of the formula (1) as mentioned in the solution hint.
$ \Rightarrow \sin \left( {\dfrac{\pi }{4} + \pi x} \right) = 0$
Step 4: On solving the expression obtained in step 3.
$ \Rightarrow \left( {\dfrac{\pi }{4} + \pi x} \right) = n\pi $
Where, $n \in Z$
$
\Rightarrow \pi x = n\pi - \dfrac{\pi }{4} \\
 \Rightarrow \pi x = \pi \left( {n - \dfrac{1}{4}} \right) \\
$
Now, eliminating $\pi $ from the both sides of the obtained expression just above.
$ \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,
$ \Rightarrow n \in (1,2,3,.........98,100)$
On substituting the value of n we can obtain the value of x hence,
$x \in \left( {\dfrac{3}{4},\dfrac{7}{4},\dfrac{{11}}{4},...........,\dfrac{{399}}{4}} \right)$

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

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 $n \in (1,2,3,............,99,100)$
To make the given trigonometric expression in form of $\sin A\cos B + \cos A\sin A$ it is necessary to multiply with $\sqrt 2 $ in the both sides of the given trigonometric expression.