Question

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

Answer 1

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 $\left[ 0,\pi \right]$.