Question

How do you solve \[\cos 2x + 5\cos x + 3 = 0\]?

Answer 1

Hint: Use trigonometric identity to convert twice of the angle to the angle. Substitute cosine of angle as a variable and transform the complete equation into a new variable. Use a method of determinant to solve for the value of the variable from the given quadratic equation. Compare the quadratic equation with general quadratic equation and substitute values in the formula of finding roots of the equation.
* \[1 + \cos 2x = 2{\cos ^2}x\]
* For a general quadratic equation \[a{x^2} + bx + c = 0\], roots are given by formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step-by-step solution:
We are given the equation \[\cos 2x + 5\cos x + 3 = 0\]
Substitute the value of \[\cos 2x = 2{\cos ^2}x - 1\]using the identity \[1 + \cos 2x = 2{\cos ^2}x\]on left hand side of the equation
\[ \Rightarrow 2{\cos ^2}x - 1 + 5\cos x + 3 = 0\]
Add the constant terms on left hand side of the equation
\[ \Rightarrow 2{\cos ^2}x + 5\cos x + 2 = 0\]
This becomes a quadratic equation in cosine
Substitute the value of \[\cos x = t\]
\[ \Rightarrow 2{t^2} + 5t + 2 = 0\]................… (1)
This becomes quadratic equation in ‘t’
We know that the general quadratic equation is \[a{x^2} + bx + c = 0\] where ‘a’, ‘b’, and ‘c’ are constant values.
On comparing the quadratic equation in equation (1) with general quadratic equation \[a{x^2} + bx + c = 0\], we get \[a = 2,b = 5,c = 2\]
Substitute the values of a, b and c in the formula of finding roots of the equation.
\[ \Rightarrow t = \dfrac{{ - (5) \pm \sqrt {{{(5)}^2} - 4 \times 2 \times 2} }}{{2 \times 2}}\]
Square the terms under the square root in numerator of the fraction
\[ \Rightarrow t = \dfrac{{ - 5 \pm \sqrt {25 - 4 \times 2 \times 2} }}{{2 \times 2}}\]
Multiply the values inside the square root in numerator of the fraction and in the denominator of the fraction
\[ \Rightarrow t = \dfrac{{ - 5 \pm \sqrt {25 - 16} }}{4}\]
\[ \Rightarrow t = \dfrac{{ - 5 \pm \sqrt 9 }}{4}\]
We can write \[\sqrt 9 = \sqrt {{3^2}} \]. Substitute this value in the numerator of the fraction.
\[ \Rightarrow t = \dfrac{{ - 5 \pm \sqrt {{3^2}} }}{4}\]
Cancel square root by square power in the numerator
\[ \Rightarrow t = \dfrac{{ - 5 \pm 3}}{4}\]
So, \[t = \dfrac{{ - 5 - 3}}{4}\]and \[t = \dfrac{{ - 5 + 3}}{4}\]
Solve the numerator value
So, \[t = \dfrac{{ - 8}}{4}\] and \[t = \dfrac{{ - 2}}{4}\]
Cancel same factors from numerator and denominator
So, \[t = - 2\] and \[t = \dfrac{{ - 1}}{2}\]
Substitute back the value of \[t = \cos x\]
So, \[\cos x = - 2\] and \[\cos x = - \dfrac{1}{2}\]
We know that the value of sine and cosine lie between -1 and 1 and -2 is less than -1, so -2 value is rejected.
\[ \Rightarrow \cos x = - \dfrac{1}{2}\]
We know cosine is negative in the second and third quadrant.

We know,\[\cos \dfrac{\pi }{3} = \dfrac{1}{2}\]
Also, we have \[\cos \left( {\pi - \theta } \right) = - \cos \theta \]
So, we can write \[\cos \left( {\pi - \dfrac{\pi }{3}} \right) = - \dfrac{1}{2}\]
i.e. \[\cos \left( {\dfrac{{2\pi }}{3}} \right) = - \dfrac{1}{2}\]
Solving for the value of x,
\[ \Rightarrow \cos x = \cos \dfrac{{2\pi }}{3}\]
Take inverse cosine function on both sides and cancel cosine function by inverse cosine
\[ \Rightarrow {\cos ^{ - 1}}\left( {\cos x} \right) = {\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }}{3}} \right)\]
\[ \Rightarrow x = \dfrac{{2\pi }}{3}\]

\[\therefore \]Solution of the equation \[\cos 2x + 5\cos x + 3 = 0\] is \[x = \dfrac{{2\pi }}{3}\]

Note: Many students leave their answer in the form of a solution of cosine of angle which is wrong as the variable in the equation is ‘x’. Keep in mind the equation is in variable ‘x’ so we have to give the value of x. Also, many students make the mistake of choosing the value of the root of cosine as -2 which is wrong.