Question

How do you find the general solution of $2{\cos ^2}x + 3\cos x - 2 = 0$?

Answer 1

Different sums have different techniques to get the solution. Usually we try to transform the trigonometric equation in some trigonometric identity which helps us most of the time but in this particular question we have been given a trigonometric equation in quadratic form. Solving this by factorization we will get our desired solution.

Complete step by step answer:
In this particular question if we observe carefully, the given trigonometric equation is in quadratic form.
We know that by solving the quadratic equation we get two values of the variable. In this case we will get the value of $x$. That is our required solution.
Let us start solving the sum by solving the above trigonometric equation which is in quadratic form.
$2{\cos ^2}x + 3\cos x - 2 = 0$
Finding the factors of the above equation we get
$2{\cos ^2}x - \cos x + 4\cos x - 2 = 0$
Solving further we get
$\cos x\left( {2\cos x - 1} \right) + 2\left( {2\cos x - 1} \right) = 0$
Taking $\left( {2\cos x - 1} \right)$ common we get
$\left( {2\cos x - 1} \right)\left( {\cos x + 2} \right) = 0$
Hence, either $\left( {2\cos x - 1} \right) = 0 - - - (1)$
Or
$\left( {\cos x + 2} \right) = 0 - - - (2)$
Solving equation (1) we get
$\left( {2\cos x - 1} \right) = 0 \\\ \Rightarrow 2\cos x = 1 \\\ \Rightarrow \cos x = \dfrac{1}{2} \\\$
And we know that when$\cos x = \dfrac{1}{2}$ value of $x = \dfrac{\pi }{3}$ and $\dfrac{{5\pi }}{3}$
Solving equation (2) we get
$\left( {\cos x + 2} \right) = 0 \\\ \Rightarrow \cos x = - 2 \\\$
But we know that value of $\cos x$ can never be -2
Hence we cannot consider equation (2)
Therefore solution to our trigonometric equation in quadratic form is
$x = \dfrac{\pi }{3}$ and $\dfrac{{5\pi }}{3}$ respectively.

Note: When given a trigonometric equation, always try to bring it to basic form using identities or try to identify the equation if it is a certain form whose process of solving is known. Generally when given a quadratic equation we can solve it by factorization process as we have done above or we can solve it using the formula $x = \dfrac{{\sqrt {{b^2} - 4ac} }}{{2a}}$ for the quadratic equation $a{x^2} + bx + c = 0$.