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Question: How do you solve \({{\csc }^{2}}x-2=0\)?...

How do you solve csc⁡2x−2=0{{\csc }^{2}}x-2=0?

Explanation

Solution

We have a trigonometric equation in the expression, we will simplify the equation to get the value or the range of the possible values of xx. We will convert the cosec function to sine function and then use the general solution of sine function to get the answer.

Complete step by step answer:
We have the given expression given to us as csc⁡2x−2=0{{\csc }^{2}}x-2=0
On rearranging the equation, we get:
⇒csc⁡2x=2\Rightarrow {{\csc }^{2}}x=2
Now on taking the square root on both the sides we get:
⇒csc⁡x=±2\Rightarrow \csc x=\pm \sqrt{2}
Now on using the trigonometric inverse function we can write the equation as:
⇒x=csc⁡−1(±2)\Rightarrow x={{\csc }^{-1}}\left( \pm \sqrt{2} \right)
Now we know about the trigonometric identity that csc⁡−1x=sin⁡−11x{{\csc }^{-1}}x={{\sin }^{-1}}\dfrac{1}{x}
On substituting the value in the equation, we get:
⇒x=sin⁡−1(±12)\Rightarrow x={{\sin }^{-1}}\left( \pm \dfrac{1}{\sqrt{2}} \right)
We know that sin⁡−1(−x)=−sin⁡−1(x){{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}\left( x \right)and since we know that the principal value of sin⁡−1(12){{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)is π4\dfrac{\pi }{4} and 3π4\dfrac{3\pi }{4} radians
The principal value of sin⁡−1(−12){{\sin }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)will be −π4-\dfrac{\pi }{4} and −3π4-\dfrac{3\pi }{4} radians
And since the value of sin⁡−1(x){{\sin }^{-1}}\left( x \right)will be the same after every 2π2\pi radians, the solution of the expression in the general format will be as:
S=\left\\{ \pm \dfrac{\pi }{4}+2\pi n,\pm \dfrac{3\pi }{4}+2\pi n \right\\}, where nn is any integer.
The set SS is the required solution.

Note: The formula used over here is for sin⁡(nπ+x)\sin (n\pi +x) ,
It is to be remembered that sin⁡(nπ+x)=(−1)nsin⁡x\sin (n\pi +x)={{(-1)}^{n}}\sin x
Basic trigonometric formulas should be remembered to solve these types of sums.
The inverse trigonometric function of sin⁡x\sin x which is sin⁡−1x{{\sin }^{-1}}x used in this sum.
For example, if sin⁡x=a\sin x=a then x=sin⁡−1ax={{\sin }^{-1}}a .
And sin⁡−1(sin⁡x)=x{{\sin }^{-1}}(\sin x)=x is a property of the inverse function.
There also exists inverse functions for the other trigonometric relations such as cos and tan.
The inverse function is used to find the angle xx from the value of the trigonometric relation.
The trigonometric function given to us in the question is csc⁡x\csc x which is actually pronounced as cosec but written as csc. It is the reciprocal function of sine.