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Question: How do you find the exact solutions of \(\cos 2x - \cos x = 0\) in the interval \(\left[ {0,2\pi } \...

How do you find the exact solutions of cos⁡2x−cos⁡x=0\cos 2x - \cos x = 0 in the interval [0,2π)\left[ {0,2\pi } \right)?

Explanation

Solution

Here we can turn the above given equation into the quadratic equation in cos⁡x\cos x and then we can find the different values of the cos⁡x\cos x and then according to its value we can find the value of xx from the graph of the cos⁡x\cos x

Complete step by step solution:
Here we need to find the value of xx but in the interval which is given as [0,2π)\left[ {0,2\pi } \right)
In this interval we must know that this open bracket means that 2π2\pi is not included in the interval and towards the left we have the closed bracket which means 00 is included in the interval. Hence we cannot take 2π2\pi as our answer.
So here we are given the equation as:
cos⁡2x−cos⁡x=0\cos 2x - \cos x = 0 −−−(1) - - - (1)
We know that cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x −−−−−(2) - - - - - (2)
Also we know that
sin⁡2x+cos⁡2x=1 sin⁡2x=1−cos⁡2x  {\sin ^2}x + {\cos ^2}x = 1 \\\ {\sin ^2}x = 1 - {\cos ^2}x \\\
Now we can substitute this value in the equation (2) and get:
cos⁡2x=cos⁡2x−(1−cos⁡2x) cos⁡2x=2cos⁡2x−1 −−−−−(3)  \cos 2x = {\cos ^2}x - (1 - {\cos ^2}x) \\\ \cos 2x = 2{\cos ^2}x - 1{\text{ }} - - - - - (3) \\\
Now substituting this value we get in equation (3) in the equation (1) we will get:
2cos⁡2x−1 −cos⁡x=0 2cos⁡2x−cos⁡x−1=0  2{\cos ^2}x - 1{\text{ }} - \cos x = 0 \\\ 2{\cos ^2}x - \cos x - 1 = 0 \\\
Now we get the quadratic equation in cos⁡x\cos x
Now we can write in above equation that (−cos⁡x)=(−2cos⁡x+cos⁡x)\left( { - \cos x} \right) = \left( { - 2\cos x + \cos x} \right)
We will get:
2cos⁡2x−2cos⁡x+cos⁡x−1=02{\cos ^2}x - 2\cos x + \cos x - 1 = 0
Simplifying it further we will get:
2cos⁡x(cos⁡x−1)+(cos⁡x−1)=02\cos x(\cos x - 1) + (\cos x - 1) = 0
(2cos⁡x+1)(cos⁡x−1)=0\left( {2\cos x + 1} \right)\left( {\cos x - 1} \right) = 0
So we can say either cos⁡x=−12 or cos⁡x=1\cos x = - \dfrac{1}{2}{\text{ or }}\cos x = 1
Now we can plot the graph of cos⁡x\cos x which is as:

Now we know that from the graph we can see:
cos⁡x=−12 or cos⁡x=1\cos x = - \dfrac{1}{2}{\text{ or }}\cos x = 1
cos⁡x=1 x=0  \cos x = 1 \\\ x = 0 \\\
cos⁡x=−12\cos x = - \dfrac{1}{2}
From the graph we can notice that:
For cos⁡x=−12\cos x = - \dfrac{1}{2}
x=2π3,4π3x = \dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3}

Hence we get the values as x=0,2π3,4π3x = 0,\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3}

Note:
If we do not know the graph we can use the properties of cosine function which says that:
cos⁡2π3=cos⁡(π2+π6)\cos \dfrac{{2\pi }}{3} = \cos \left( {\dfrac{\pi }{2} + \dfrac{\pi }{6}} \right)
Now we know that cos⁡(π2+x)=−sin⁡x\cos \left( {\dfrac{\pi }{2} + x} \right) = - \sin x
So we get cos⁡2π3=cos⁡(π2+π6)=−sin⁡π6=−12\cos \dfrac{{2\pi }}{3} = \cos \left( {\dfrac{\pi }{2} + \dfrac{\pi }{6}} \right) = - \sin \dfrac{\pi }{6} = - \dfrac{1}{2}
Hence we must know the properties of all the trigonometric functions in order to solve such problems.