Question
Question: How do you solve \({{\sin }^{2}}x-1=0\) and find all exact general solutions?...
How do you solve sin2x−1=0 and find all exact general solutions?
Solution
First of all we have to solve the given expression and then find the range of function and after that we have to use arithmetic progression to solve the expression to get the general solution.
Complete step by step answer:
Given that
sin2x−1=0
We know that
a2−b2=(a+b)(a−b)
Therefore, the given expression may be written as
sin2x−1=(1+sinx)(sinx−1)
Therefore,
(1+sinx)(sinx−1)=0
Now solve the above expression in parts
(1+sinx)=0
⇒sinx=−1
From above expression we get
x=23π,27π,....
And
sinx−1=0
⇒sinx=1
From above expression we get
x=2π,25π,....
Now we get
x=2π,23π,25π,...
Now to final answer we have to use the arithmetic progression
Here,
a=2π
And
d=π
We know that the formula of arithmetic progression is as follows:
Tn=2π+(n−1)π
Further solving the above expression we get
Tn=(21+n−1)π
Further solving
Tn=(n−21)π
And
Tn=(2n−1)2π
Therefore, the exact general solution of the given problem is
Tn=(n−21)π
And
Tn=(2n−1)2π
Additional information:
In these questions we have to find the range of the given function by the trigonometric method in which the function lies. After finding all the ranges we can find the exact solutions of the problem.
Note:
In such a type of question it is very important to understand the question about what the question is asking. After understanding this it is really important for us to find out the range of the function correctly. Because if we calculate the wrong range of the function the whole solution goes wrong. After that we have to combine the ranges by arithmetic progression method to get the exact solution. So it is also important to remember the formula of arithmetic progression.