Question

How do you solve for the exact solutions in the interval $\left[ 0,2\pi \right]$ of $\cos 2x-\cos x=0?$

Answer 1

Hint: We will use some familiar identities to solve the given equation. We should remember the identity $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x.$ We will also use the identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1.$ We will do some rearrangements in the equation after applying the necessary identities to find the solution.

Complete step by step solution:
Let us consider the given trigonometric equation $\cos 2x-\cos x=0.$
We need to find the exact solutions of the given function in the given interval $\left[ 0,2\pi \right].$
Let us use the identity $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x.$
When we apply this equation in the given equation, we will get ${{\cos }^{2}}x-{{\sin }^{2}}x-\cos x=0.$
Let us recall another trigonometric identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1.$
If we rearrange this identity, we will get $1-{{\cos }^{2}}x={{\sin }^{2}}x.$
Let us substitute this in the obtained equation.
We will get ${{\cos }^{2}}x-\left( 1-{{\cos }^{2}}x \right)-\cos x=0.$
Let us open the bracket to get ${{\cos }^{2}}x-1+{{\cos }^{2}}x-\cos x=0.$
We will get $2{{\cos }^{2}}x-\cos x-1=0$
We have obtained a trigonometric equation that resembles a quadratic equation.
Let us put $\cos x=a$ in the above equation to convert it into a quadratic equation.
We will get $2{{a}^{2}}-a-1=0.$
Let us use the quadratic formula to find the solution of the above quadratic equation.
The quadratic formula is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for an equation of the form $a{{x}^{2}}+bx+c=0.$
When we will apply this in our problem, we will get $a=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 2\times \left( -1 \right)}}{2\times 2}.$
From this we will obtain, $a=\dfrac{1\pm \sqrt{1+8}}{4}=\dfrac{1\pm \sqrt{9}}{4}=\dfrac{1\pm 3}{4}.$
So, we will get $a=\dfrac{1+3}{4}=\dfrac{4}{4}=1$ or $a=\dfrac{1-3}{4}=\dfrac{-2}{4}=\dfrac{-1}{2}.$
Since $\cos x=a,$ we will get $\cos x=1$ or $\cos x=\dfrac{-1}{2}.$
This is possible when $x=0$ or $x=\pi \pm \dfrac{\pi }{3}.$ Because the Cosine function is negative in the second and third quadrants.

Hence the solution is $x=0,\dfrac{2\pi }{3},\dfrac{4\pi }{3}.$

Note: Remember the identity ${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}.$ From this, we will get $\cos 2x=2{{\cos }^{2}}x-1.$
We can arrive at the quadratic equation by using this identity also. If we put this identity in the given equation, we will get $2{{\cos }^{2}}x-1-\cos x=0.$ And the rest of the procedure is the same.