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Question: How do you solve \( \cos 2x = \cos x \) ?...

How do you solve cos2x=cosx\cos 2x = \cos x ?

Explanation

Solution

In order to determine the solution of the above trigonometric equation, Rewrite the equation using trigonometric identity cos2x=2cos2x1\cos 2x = 2{\cos ^2}x - 1 . You will obtain a quadratic equation, compare it with the standard equation to get the value of variables a,b,ca,b,c . Since a+b+c=0a + b + c = 0 , obtain the solution by putting cosx=0\cos x = 0 and cosx=ca=12\cos x = \dfrac{c}{a} = - \dfrac{1}{2} .

Complete step by step answer:
We are given a trigonometric equation cos2x=cosx\cos 2x = \cos x and we have to find its solution
Since in the question we have not given any constraint for the value of xx , so we will be finding the solution from 00 to 2π2\pi
We are going to rewrite our using the identity of trigonometry cos2x=2cos2x1\cos 2x = 2{\cos ^2}x - 1
2cos2x1=cosx 2cos2xcosx1=0  2{\cos ^2}x - 1 = \cos x \\\ 2{\cos ^2}x - \cos x - 1 = 0 \\\
Let X=cosxX = \cos x
As we can see we obtained a quadratic equation, lets compare the above equation with the standard quadratic equation aX2+bX+c=0a{X^2} + bX + c = 0 , we get
a=2 b=1 c=1  a = 2 \\\ b = - 1 \\\ c = - 1 \\\
Since a+b+c=211=0a + b + c = 2 - 1 - 1 = 0 , we can say that there are two real roots as
a. cosx=0\cos x = 0 , x=cos1(0)\Rightarrow x = {\cos ^{ - 1}}\left( 0 \right) x=0or2π\Rightarrow x = 0\,or\,2\pi
b. cosx=ca=12\cos x = \dfrac{c}{a} = - \dfrac{1}{2} , x=cos1(12)x=2π3or4π3\Rightarrow x = {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) \Rightarrow x = \dfrac{{2\pi }}{3}\,or\,\dfrac{{4\pi }}{3}

Therefore, the solution of given trigonometric equation is x=0,2π3,4π3,2πx = 0,\dfrac{{2\pi }}{3},\,\,\dfrac{{4\pi }}{3},2\pi between 00 to 2π2\pi .

Additional Information:
1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
2. Even Function – A function f(x)f(x) is said to be an even function ,if f(x)=f(x)f( - x) = f(x) for all x in its domain.
Odd Function – A function f(x)f(x) is said to be an even function ,if f(x)=f(x)f( - x) = - f(x) for all x in its domain.
We know that sin(θ)=sinθ.cos(θ)=cosθandtan(θ)=tanθ\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta
Therefore, sinθ\sin \theta and tanθ\tan \theta their reciprocals, cosecθ\cos ec\theta and cotθ\cot \theta are odd functions whereas cosθ\cos \theta and its reciprocal secθ\sec \theta are even functions.
3. Periodic Function= A function f(x)f(x) is said to be a periodic function if there exists a real number T > 0 such that f(x+T)=f(x)f(x + T) = f(x) for all x.

Note: 1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2. Period of cosine function is 2π2\pi .
3. The domain of cosine function is in the interval [0,π]\left[ {0,\pi } \right] and the range is in the interval [1,1]\left[ { - 1,1} \right] .