Question

How do you solve: \[2{{\cos }^{-1}}x=\pi \]?

Answer 1

Hint: Divide both the sides of the given expression with 2 to simplify. Now, take cosine function on both the sides and use the formula: - \[\cos \left( {{\cos }^{-1}}x \right)=x\], where \[-1\le x\le 1\], to simplify the L.H.S. In the R.H.S. use the formula or value \[\cos \left( \dfrac{\pi }{2} \right)=0\] to get the answer.

Complete step by step solution:
Here, we have been provided with the inverse trigonometric equation containing the inverse cosine function given as: - \[2{{\cos }^{-1}}x=\pi \]. We are asked to find the values of x.
\[\because 2{{\cos }^{-1}}x=\pi \]
Dividing both the sides with 2, we get,
\[\Rightarrow {{\cos }^{-1}}x=\dfrac{\pi }{2}\]
Now, taking cosine function on both the sides, we get,
\[\Rightarrow \cos \left( {{\cos }^{-1}}x \right)=\cos \left( \dfrac{\pi }{2} \right)\]
We know that \[\cos \left( {{\cos }^{-1}}x \right)=x\] when we have \[-1\le x\le 1\] which is also the range of sine and cosine function, so we have,
\[\Rightarrow x=\cos \left( \dfrac{\pi }{2} \right)\]
We know that the value of an odd multiple of \[\dfrac{\pi }{2}\] for a cosine function is 0. In the R.H.S. of the above expression we have \[\left( \dfrac{\pi }{2} \right)\] which can be written as: - \[\left( 1\times \dfrac{\pi }{2} \right)\]. Clearly, 1 is an odd number, so we have an odd multiple of \[\left( \dfrac{\pi }{2} \right)\] as the argument of the cosine function. Therefore, using the formula: - \[\cos \left[ \left( 2n+1 \right)\dfrac{\pi }{2} \right]=0\], where \[n\in \] integers, we get,
\[\Rightarrow x=0\]
Hence, the value of x is 0.

Note: One can also solve the given equation by drawing the graph of either inverse cosine function or cosine function. In the graph of a cosine function you can clearly see that at \[x=\dfrac{\pi }{2}\] the value of \[\cos x=0\]. There can be a different approach also using which we can solve the equation. After dividing both the sides with 2 we will use the identity \[{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}\] and substitute the value of \[{{\cos }^{-1}}x\] in the given equation. Using this identity, we will get the relation: - \[{{\sin }^{-1}}x=0\]. Now, we know that the range of \[{{\sin }^{-1}}x\] is \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\], so using this we can say that the only value at which \[{{\sin }^{-1}}x\] is 0 will be the angle 0 degrees or 0 radian. You must remember the range of all inverse trigonometric functions because we have many angles for which the value of cosine function is 0 but we need to consider only the principal value that lies in the range \[\left[ 0,\pi \right]\].