Solveeit Logo

Question

Question: What is the inverse cosine of 2?...

What is the inverse cosine of 2?

Explanation

Solution

Hint : We first find the principal value of x for which cos1(2){{\cos }^{-1}}\left( 2 \right). In that domain, equal value of the same ratio gives equal angles. We find the angle value for x. At the end we also find the general solution for the equation cos(x)=2\cos \left( x \right)=2.

Complete step by step solution:
We find the value of cos1(2){{\cos }^{-1}}\left( 2 \right). We need to find x for which cos(x)=2\cos \left( x \right)=2.
We know that in the principal domain or the periodic value of π2xπ2-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2} for cos(x)\cos \left( x \right), if we get cosa=cosb\cos a=\cos b where π2a,bπ2-\dfrac{\pi }{2}\le a,b\le \dfrac{\pi }{2} then a=ba=b.
But the range for cos(x)\cos \left( x \right) is 1cos(x)1-1\le \cos \left( x \right)\le 1. So, it is not possible to find xx for which cos(x)=2\cos \left( x \right)=2.
Now we prove that graphically.
We also can show the solutions (primary and general) of the equation cos(x)=2\cos \left( x \right)=2 through the graph. We take y=cos(x)=2y=\cos \left( x \right)=2. We got two equations y=cos(x)y=\cos \left( x \right) and y=2y=2. We place them on the graph and find the solutions as their intersecting points.

We can see that there is no intersecting point for the curves and the equation has no solution.

Note : Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to x-\infty \le x\le \infty . In that case we have to use the formula x=nπ±ax=n\pi \pm a for cosa=cosb\cos a=\cos b where π2aπ2-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}.