Question
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. Also find all the solutions in [0, 100].
Solution
- According to given in the question we have to solve the given trigonometric expression sinπx+cosπx=0. in between [0, 100] so, first of all we will try to make the given trigonometric expression in form of sinAcosB+cosAsinA so that we can apply the formula as given below:
sin(A+B)=sinAcosB+cosAsinA...................(1)
So, to convert the given trigonometric expression sinπx+cosπx=0 in form of sinAcosB+cosAsinA we have to multiply and divide with 2 in the numerator and denominator of the expression and then we have to convert 2 into sin and cos with the help of the formula given below:
sin4π=2...................(2)
cos4π=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 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 in the both sides of the given trigonometric expression to obtain the expression in form of sinAcosB+cosAsinA
Hence, on multiplying and divide with 2 in the numerator and denominator of the given expression,
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.
⇒sin4πcosπx+cos4πsinπx=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.
⇒sin(4π+πx)=0
Step 4: On solving the expression obtained in step 3.
⇒(4π+πx)=nπ
Where, n∈Z
⇒πx=nπ−4π ⇒πx=π(n−41)
Now, eliminating π from the both sides of the obtained expression just above.
⇒x=(n−41)
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)
On substituting the value of n we can obtain the value of x hence,
x∈(43,47,411,...........,4399)
Hence with the help of the formula (1), (2) and (3) we have obtained the value of the given trigonometric expression sinπx+cosπx=0 is x=(n−41) and range from [0, 100] is x∈(43,47,411,...........,4399)
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)
To make the given trigonometric expression in form of sinAcosB+cosAsinA it is necessary to multiply with 2 in the both sides of the given trigonometric expression.