Question

Find the general solution:
$\cos 3x + \cos x - \cos 2x = 0$

Answer 1

We can start the problem with analyzing $cos\left( {3x} \right) = cos\left( {x{\text{ }} + {\text{ }}2x} \right)$. Then using that formula $\cos (a + b) = \cos a\cos b - \sin a\sin b$ we can simplify and then, we substitute the value in$\cos 3x + \cos x - \cos 2x = 0$, and then on further simplification, we get the solution.

Complete step by step Answer:

We are to find a general solution of, $\cos 3x + \cos x - \cos 2x = 0$
we start with,
$cos\left( {3x} \right) = cos\left( {x{\text{ }} + {\text{ }}2x} \right)$
Now, as, $\cos (a + b) = \cos a\cos b - \sin a\sin b$
$= {\text{ }}cos\left( x \right)cos\left( {2x} \right){\text{ }} - {\text{ }}sin\left( x \right)sin\left( {2x} \right)$
As, $\cos 2x = 2{\cos ^2}x - 1$and $\sin 2x = 2\sin x\cos x$, we get,
$= cos(x)(2co{s^2}(x) - 1) - 2si{n^2}(x)cos(x)$
Now, on simplifying, and using $1 - {\cos ^2}x = {\sin ^2}x$ we get,
$= 2co{s^3}(x) - cos(x) - 2cos(x)(1 - co{s^2}(x))$
On adding like terms we get,
$= 4co{s^3}(x) - 3cos(x)$
Therefore $cos\left( {3x} \right){\text{ }} + {\text{ }}cos\left( x \right){\text{ }} - {\text{ }}cos\left( {2x} \right){\text{ }} = {\text{ }}0$can also be written as
$4co{s^3}(x) - 3cos(x) + cos(x) - 2co{s^2}(x) + 1 = 0$
On adding the like terms we get,
$\Rightarrow 4co{s^3}(x) - 2co{s^2}(x) - 2cos(x) + 1 = 0$
On taking terms common we get,
$\Rightarrow 2{\cos ^2}x(2cos(x) - 1) - 1(2cos(x) - 1) = 0$
$\Rightarrow (2co{s^2}(x) - 1)(2cos(x) - 1) = 0$
Thus we can say that $2co{s^2}(x) = 1$or $2cos\left( x \right){\text{ }} = {\text{ }}1$
Hence $cos\left( x \right){\text{ }} = {\text{ }}\dfrac{1}{{\sqrt 2 }}$or $cos\left( x \right){\text{ }} = {\text{ - }}\dfrac{1}{{\sqrt 2 }}$or $cos\left( x \right){\text{ }} = {\text{ }}\dfrac{1}{2}$
Hence $x{\text{ }} = {\text{ }}45^\circ {\text{ }}or{\text{ }}135^\circ {\text{ }}or{\text{ }}225^\circ {\text{ }}or{\text{ }}315$°
OR $x{\text{ }} = {\text{ }}60^\circ {\text{ }}or{\text{ }}300^\circ$.

Note: The general solution of the equation should be stated as the solution in the range of $- 360^\circ$to $360^\circ$. We will consider the values as general values if they are inside that given range. Otherwise, the decision is not general. We should consider all the possible cases and solve for all of them.