Question

How do you solve ${{\cos }^{3}}x+{{\cos }^{2}}x-\cos x=1$ for $0\le x\le 2\pi ?$

Answer 1

Hint: We will use the concept of trigonometric identities to solve the above question. To find the value of x we will first take 1 from the right of the equation to the left of the equation. Then, we will use the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and write ${{\cos }^{2}}x-1=-{{\sin }^{2}}x$ and we will take $\cos x$ common from ${{\cos }^{3}}x$ and $-\cos x$and write again ${{\cos }^{2}}x-1=-{{\sin }^{2}}x$, and then we will use trigonometric equation to get value of x.

Complete step-by-step solution:
We will use the concept of trigonometric identities to solve the above equation. First, we will take 1 from RHS to the LHS.
$\Rightarrow {{\cos }^{3}}x+{{\cos }^{2}}x-\cos x=1$
$\Rightarrow {{\cos }^{3}}x+{{\cos }^{2}}x-\cos x-1=0$
Now, we will take $\cos x$ common from ${{\cos }^{3}}x$ and $-\cos x$, then we will get:
\[\Rightarrow \cos x\left( {{\cos }^{2}}x-1 \right)+{{\cos }^{2}}x-1=0\]
Now, from trigonometric identities we know that $\left[ 0,2\pi \right]$ ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, so we can write ${{\cos }^{2}}x-1=-{{\sin }^{2}}x$.
\[\Rightarrow \cos x\left( -{{\sin }^{2}}x \right)-{{\sin }^{2}}x=0\]
Now, after taking $-{{\sin }^{2}}x$ common we will get:
\[\Rightarrow -{{\sin }^{2}}x\left( \cos x+1 \right)=0\]
\[\Rightarrow {{\sin }^{2}}x\left( \cos x+1 \right)=0\]
Now, we will equate each ${{\sin }^{2}}x$ and cos x + 1 both equal to 0.
\[\therefore {{\sin }^{2}}x=0\] and $\cos x+1=0$
\[\Rightarrow \sin x=0\] and $\cos x=-1$
Now, from trigonometric equation we know that when
 $\begin{align}
  & \sin x=0 \\
 & \Rightarrow x=n\pi \\
\end{align}$ , where n belongs to integer.
Now, when $\cos x=-1$
$\Rightarrow x=2n\pi \pm \pi $ , here also n belong to integers.
But, from question we know that x belongs to $0\text{ to }2\pi $
So, we have to find all such x which belong to $0\text{ to }2\pi $.
Now, we will put a different value of n in the general solution and obtain all the values x.
So, for $x=n\pi $, when n = 0 we have x = 0,
When n = 1, we have $x=\pi $, and when n = 2, we have $x=2\pi $.
Similarly, for $x=2n\pi \pm \pi $, when we have n = 0, $x=\pi $, it is only value for $x=2n\pi \pm \pi $ which lies between $0\text{ to }2\pi $.
So, the value of x which satisfies ${{\cos }^{3}}x+{{\cos }^{2}}x-\cos x=1$, is $0,\pi ,2\pi $.
This is our required solution.

Note: Students are required to note that when we have to solve a trigonometric equation and we are also given the domain of x then we have to all the values of x in the domain which satisfies the given trigonometric equation. Also, note that in trigonometry value repeats after a certain period.