Question

How do you solve for x by satisfying the given trigonometric equation $\cos 2x+2\cos x+1=0$?

Answer 1

We start solving the problem by making use of the result $\cos 2x=2{{\cos }^{2}}x-1$ in the given trigonometric equation. We then factorize the obtained quadratic equation in cosine function and then equate the obtained factors to 0 to get the values of $\cos x$ which satisfies the given trigonometric equation. We then make use of the facts that the general solution of $\cos x=0$ is $\left( 2n+1 \right)\dfrac{\pi }{2}$, $n\in Z$ and the general solution of $\cos x=-1$ is $\left( 2n+1 \right)\pi$, $n\in Z$ to get the required solution of the given trigonometric equation.

Complete step by step answer:
According to the problem, we are asked to find the values of x satisfying the given trigonometric equation $\cos 2x+2\cos x+1=0$.
We have given the trigonometric equation $\cos 2x+2\cos x+1=0$ ---(1).
We know that $\cos 2x=2{{\cos }^{2}}x-1$. Let us use this result in equation (1).
$\Rightarrow 2{{\cos }^{2}}x-1+2\cos x+1=0$.
$\Rightarrow 2{{\cos }^{2}}x+2\cos x=0$.
$\Rightarrow 2\cos x\left( \cos x+1 \right)=0$.
$\Rightarrow 2\cos x=0$, $\cos x+1=0$.
$\Rightarrow \cos x=0$, $\cos x=-1$ ---(2).
We know that the general solution of $\cos x=0$ is $\left( 2n+1 \right)\dfrac{\pi }{2}$, $n\in Z$ and the general solution of $\cos x=-1$ is $\left( 2n+1 \right)\pi$, $n\in Z$.
$\Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{2}$, $\left( 2n+1 \right)\pi$ for $n\in Z$.
So, we have found the solution set of the given trigonometric equation $\cos 2x+2\cos x+1=0$ as $x=\left( 2n+1 \right)\dfrac{\pi }{2}$, $\left( 2n+1 \right)\pi$ for $n\in Z$.
$\therefore$ The solution set of the given trigonometric equation $\cos 2x+2\cos x+1=0$ is $x=\left( 2n+1 \right)\dfrac{\pi }{2}$, $\left( 2n+1 \right)\pi$ for $n\in Z$.

Note: We should perform each step carefully in order to avoid confusions and calculation mistakes. Here we need to find the general solution for the given trigonometric equation to solve the given problem. We can also find the principal solution for the given trigonometric solution which can also be found by substituting $n=0$ in the general solution. Similarly, we can expect problems to find the general solution of the trigonometric equation ${{\cos }^{2}}x-\cos x-2=0$.