Question

How do you solve $\cos 2(x) - \cos (x) - 1 = 0$?

Answer 1

Hint: In this question, the given trigonometric equation needs to be solved to obtain the value for $\cos x$ and then for the variable x. Firstly, we need to simplify the first term. We do this using a trigonometric formula for $\cos 2x$. The formula is given as, $\cos 2x = 2{\cos ^2}x - 1$. Then, substitute the obtained expression in the given equation and we get a quadratic equation. We solve this quadratic equation and obtain the roots which gives us the required result.

Complete step by step solution:
Given the trigonometric equation $\cos 2(x) - \cos (x) - 1 = 0$ …… (1)
We are asked to solve the above expression given in the equation (1).
Firstly, we will try to simplify the first term and then substitute back in the original equation and simplify it further.
So let us consider the first term given by $\cos 2(x)$
Which can also be written as $\cos 2x$.
We have different formulas for $\cos 2x$. We make use of one among them to solve the problem.
Since the original equation is in terms of cosine, we use the formula for $\cos 2x$ given by,
$\cos 2x = 2{\cos ^2}x - 1$
Substitute this expression in the equation (1), we get,
$ \Rightarrow 2{\cos ^2}x - 1 - \cos (x) - 1 = 0$
Rearranging the terms we get,
$ \Rightarrow 2{\cos ^2}x - \cos (x) - 1 - 1 = 0$
$ \Rightarrow 2{\cos ^2}x - \cos x - 2 = 0$
Now we take $t = \cos x$. So that it will be easier to solve. Hence we have,
$ \Rightarrow 2{t^2} - t - 2 = 0$ …… (2)
Note that the above equation is a quadratic equation.
If $a{x^2} + bx + c = 0$ is a quadratic equation, then we find the roots using quadratic formula given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Note that here $a = 2$, $b = - 1$ and $c = - 2$.
Hence we get,
$t = \dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4 \times 2 \times ( - 2)} }}{{2 \times 2}}$
$ \Rightarrow t = \dfrac{{1 \pm \sqrt {1 + 16} }}{4}$
$ \Rightarrow t = \dfrac{{1 \pm \sqrt {17} }}{4}$
Now substituting back $t = \cos x$ we get,
$ \Rightarrow \cos x = \dfrac{{1 \pm \sqrt {17} }}{4}$
When $\cos x = \dfrac{{1 + \sqrt {17} }}{4}$
$ \Rightarrow \cos x = \dfrac{{1 + 4.12}}{4}$
$ \Rightarrow \cos x = \dfrac{{5.12}}{4}$
$ \Rightarrow \cos x = 1.28$, which is not possible.
Since the range of cosine is $ - 1 \leqslant \cos x \leqslant 1$.
When $\cos x = \dfrac{{1 - \sqrt {17} }}{4}$
$ \Rightarrow \cos x = \dfrac{{1 - 4.12}}{4}$
$ \Rightarrow \cos x = \dfrac{{ - 3.12}}{4}$
$ \Rightarrow \cos x = - 0.78$
$ \Rightarrow x = {\cos ^{ - 1}}( - 0.78)$
$ \Rightarrow 141.26$
i.e. $x = {141^\circ }26$

Hence the solution for the expression $\cos 2(x) - \cos (x) - 1 = 0$ is given by $x = {141^\circ }26$.

Note :
We can check whether the obtained answer is correct or not, by substituting back in the original expression. If the equation satisfies the value obtained, then our answer is the required one.
Students must remember the formulas related to sine and cosine.
Some of them are given below.
(1) $\cos 2x = {\cos ^2}x - {\sin ^2}x$
(2) $\cos 2x = 2{\cos ^2}x - 1$
(3) $\cos 2x = 1 - 2{\sin ^2}x$
(4) $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$
(5) $\sin 2x = 2\sin x\cos x$