Question
Question: How do you solve \(2{{\cos }^{2}}x-5\cos x+2=0?\)...
How do you solve 2cos2x−5cosx+2=0?
Solution
We will convert the given trigonometric equation to a quadratic equation by equating cosx with a variable c. Then we will solve the quadratic equation for c using the quadratic formula.
Complete step by step solution:
Let us consider the given trigonometric equation 2cos2x−5cosx+2=0.
We are asked to solve the given trigonometric equation.
Let us first convert the given trigonometric function into a quadratic equation in c by equating cosx into c.
The trigonometric equation will become a quadratic equation 2c2−5c+2=0.
We can use the quadratic formula to solve the obtained quadratic equation.
Suppose that we have a quadratic equation ax2+bx+c=0. Then we can use the following formula which is called the quadratic formula to solve the equation for x:x=2a−b±b2−4ac.
Let us compare the coefficients to get a=2,b=−5 and c=2.
So, the quadratic formula for our equation is c=2×2−(−5)±(−5)2−4×2×2.
We can write this equation as c=45±25−16.
That can be written as c=45±9.
When we write this equation by substituting 3=9, we will get c=45±3.
From this, we will get the possible values of the variable c as c=45−3 or c=45+3.
And this will give us c=42=21 or c=48=2.
We know that cosx=c. So, we will get cosx=2 or cosx=21.
But we know that the range of the Cosine function is [−1,1].
So, it is not possible that cosx=2.
Therefore, we will get cosx=21.
We know that the values of x for which cosx=21 are ±3π.
In general, x=±3π+2nπ,n∈Z.
Hence the solution of the given equation is given by x=±3π+2nπ,n∈Z.
Note: We can factorize the obtained quadratic equation 2c2−5c+2=0 as (c−2)(2c−1)=0. Then from the first factor, we will get c=2 and from the second fraction, we will get 2c=1⇒c=21.