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Question

Question: How do you solve \({\cos ^2}x - \cos x = 0\) ?...

How do you solve cos2xcosx=0{\cos ^2}x - \cos x = 0 ?

Explanation

Solution

In the question, a polynomial equation of one variable is given, it is a function of cosx\cos x . To make it easier to solve the given polynomial equation, we replace cosx\cos x in the given equation with any other variable. We know that the degree of a polynomial equation is the highest power of the variable used in the polynomial and also the number of roots of a polynomial equation is equal to its degree, so several methods like factorizing the equation or by a special formula called quadratic formula or by completing the square method or quadratic formula can be used to solve the given equation; we use other methods if we are not able to factorize the equation.

Complete step by step answer:
We have to solve the equation cos2xcosx=0{\cos ^2}x - \cos x = 0
Taking cosx=t\cos x = t, we get –
t2t=0\Rightarrow {t^2} - t = 0
Now we see that this equation can be solved by simply taking t as common –
t(t1)=0 t=0,t1=0 t=0,t=1  \Rightarrow t(t - 1) = 0 \\\ \Rightarrow t = 0,\,t - 1 = 0 \\\ \Rightarrow t = 0,\,t = 1 \\\
Putting the original value of t as cosx\cos x , we get –
cosx=0,cosx=1\Rightarrow \cos x = 0,\,\cos x = 1
We know that cos0=0,cos90=1\cos 0 = 0,\,\cos 90^\circ = 1, so we get –
cosx=cos0,cosx=cos90 x=0,x=90  \Rightarrow \cos x = \cos 0,\,\cos x = \cos 90^\circ \\\ \Rightarrow x = 0^\circ ,\,x = 90^\circ \\\
Hence, the solution of the equation cos2xcosx=0{\cos ^2}x - \cos x = 0 is x=0x = 0^\circ or x=90x = 90^\circ .

Note: In this question, we have just written the general solution, there can be infinitely many solutions to the questions like this. The general solution lies in the interval [0,2π)[0,2\pi ) that is the value of x can be greater than or equal to zero but smaller than 2π2\pi so we take only these two values as the answer.