Question

How do you solve $2{{\cos }^{2}}x-3\cos x+1=0$ and find all solutions in the interval $\left( 0,2\pi \right)$?

Answer 1

Hint: We will solve the question by making it as a quadratic equation by substituting the value of $\cos x$ as $t$ and use the property of the coefficient of quadratic equations which states that when the sum of the coefficients is $0$, the roots can be found out from the value of the coefficients and then we will solve the question for the values on the given interval $\left( 0,2\pi \right)$.

Complete step by step answer:
We have the given equation as $2{{\cos }^{2}}x-3\cos x+1=0$
Since the equation in the form of a quadratic equation $a{{x}^{2}}+bx+c=0$
where $t=\cos (x)\to (1)$
We get the coefficients $a,b$ and $c$ as follows:
$a=2$
$b=-3$
$c=1$
Since in any given quadratic equation which has the sum of coefficients $0$ which means
If $a+b+c=0$, the roots are ${{x}_{1}}=1$ and ${{x}_{2}}=\dfrac{c}{a}$
Therefore, in the given equation the roots are:
 ${{x}_{1}}=1$ and ${{x}_{2}}=\dfrac{1}{2}$
Now the value of $t$ from equation $(1)$ is $\cos x$
Therefore, we can write:
$\cos x=1$ and $\cos x=\dfrac{1}{2}$
Now $\cos x=1$ when $t=0$ or $t=2\pi $ and $\cos x=\dfrac{1}{2}$ when $t=\dfrac{\pi }{3}$ or $t=\dfrac{5\pi }{3}$
But since in the question it is mentioned that we have to write the solution which is in the range $\left( 0,2\pi \right)$, the solution will be:
$\dfrac{\pi }{3},\dfrac{5\pi }{3}$, which is the required final answer.

Note: The formula for the roots of the equation when the sum of the root’s is $0$ should be remembered for solving these types of questions.
The trigonometric table should be remembered to get the appropriate value of the angle, the angle can be written as degrees or radians.
It is to be remembered which trigonometric functions are positive and negative in what quadrants to get the sign of the angle.
We have used the property of coterminal angles which means that the angles which have the same sides in the triangle.
The general form of writing coterminal angles is $\theta \pm k(2\pi )$ where $\theta $ is the initial angle, $k$ is a constant multiplied by $2\pi $ because whenever angle is multiplied by $2\pi $ it has the same sides as the initial angle.